Explicit expression of Gauduchon metric forms on non-Kähler surfaces

Shigetaka Tatsumi

0 Introduction

On a complex surface, a Hermitian metric whose associated (1.1)-form ω satisfying dd^{c}ω = 0 is called a Gauduchon metric. It is known by Gauduchon [G1] that every compact complex surface admits many Gauduchon metrics. Explicit expression of Gauduchon metrics, however, does not seem to be well-known. The purpose of this paper is to give explicit expression of Gauduchon metric forms on some VI_{o} surfaces, and VII_{o} surfaces with vanishing second Betti numbers.

All VI_{o} surfaces and VII_{o} elliptic surfaces have a finite unramified covering which are submanifolds of a higher dimensional primary Hopf manifold with a complex torus action. For these surfaces satisfying some additional conditions, we shall give a method of constructing Gauduchon metrics (Theorem 2.1).

Non-elliptic primary Hopf surfaces are treated separately. We shall give explicit expressions of Gauduchon metrics on primary Hopf surfaces which is a quotient of ℂ^{2} \ {0} by linear group actions (Theorems 4.1, 4.2).

It is also given an explicit expression of Gauduchon metrics on Inoue surfaces with b_{2} = 0. We owe much to [B] and [Tr] in this case (Theorems 5.5, 5.7).

1 Notation and preliminaries

1.1 Gauduchon metrics

We shall give the definition of Gauduchon metrics and Gauduchon metric forms.

Definition 1.1. A Hermitian metric on a complex manifold W is given in terms of local coordinates (z_{ν}) by:

ds^{2} = ∑_{α,β=1}^{n} g_{αβ̄}(z)dz_{α} ⊗ dz̄_{β},

where g_{αβ̄} is a C^{∞}-section of T^{∗} ⊗ T̄^{∗} such that:

1. \overline{g_{αβ̄}(z)} = g_{βᾱ(z)} (Hermitian symmetric).
2. ∑_{α,β=1}^{n} g_{αβ̄}(z)ζ_{α}ζ̄_{β} ≥ 0 and the equality holds if and only if ζ = 0 (positive definite).

Associated with ds^{2}, we consider the differential form of type (1,1)

ω = \sqrt{−1} ∑_{α,β=1}^{n} g_{αβ̄}(z)dz_{α} ∧ dz̄_{β},

which is called a Hermitian metric form.

Definition 1.2. Let W be a complex manifold of dimension n. A Hermitian metric ds^{2} on W is called a Gauduchon metric if its Hermitian metric form ω satisfies dd^{c}ω^{n−1} = 0, and ω is called a Gauduchon metric form. If n = 2, ds^{2} is a Gauduchon metric if and only if the associated form ω satisfies dd^{c}ω = 0.

1.2 Weighted homogeneous polynomials

We shall give a definition of weighted homogeneous polynomials.

Definition 1.3. Let m_{1}, ⋯, m_{n} be positive integers, and N a non-negative integer. A polynomial f = f(z_{1}, ⋯, z_{n}) is said to be weighted homogeneous of weight m = (m_{1}, ⋯, m_{n}) with degree N, if f is can be expressed as a linear combination of monomials z_{1}^{i_{1}}, …, z_{n}^{i_{n}} for which

m_{1}i_{i} + ⋯ + m_{n}i_{n} = N.

This is equivalent to saying that f satisfies the identity

f(t^{m_{1}}z_{1}, …, t^{m_{n}}z_{n}) = t^{N}f(z_{1}, …, z_{n}).

for any t ∈ ℂ.

2 Surfaces in a higher dimensional Hopf manifold

By a Hopf manifold of dimension n ≥ 2, we shall mean a compact complex manifold whose universal covering manifold is biholomorphic to the domain C^{n} \ {0}, where {0} denotes the origin of C^{n} ([Ko1]). By a primary Hopf manifold, we shall mean a Hopf manifold whose fundamental group is an infinite cyclic. It is known that every Hopf manifolds are submanifold of a higher dimensional primary Hopf manifold and that any VI_{o} surface or VII_{o}-elliptic surface can be embedded in to a higher dimensional Hopf manifold [Ka]. If S is replaced with its finite unramified covering, we can get a little more information on the embedding. Let S be a surface of Class VI_{o}, or an elliptic surface of VII_{o}. Note that surfaces of Class VI_{o} are always elliptic surfaces. Hence S is of algebraic dimension 1. Therefore by [Ka, Theorem 4], there is an finite unramified covering S′ of S which can be embedded holomorphically into a higher dimensional primary Hopf manifold, which is constructed as follows.

Fix a constant α ∈ C such that 0 < |α| < 1, and m = (m_{1}, …, m_{n}) ∈ Z_{>}^{n}. Let

H_{(α,m)}^{n} = W/⟨g⟩,

where

W = C^{n} \ {0},
g(z_{1}, …, z_{n}) = (α^{m_{1}}z_{1}, …, α^{m_{n}}z_{n}).

Then H_{(α,m)}^{n} is a compact complex manifold diffeomorphic to S^{1} × S^{2n−1}.

Now we shall make the first assumption.

Assumption 1. S be a VI_{o} or VII_{o}-elliptic surface embedded in H_{(α,m)}^{n}.

Let ϖ : W → H_{(α,m)}^{n} be the canonical projection. Then S̃ = ϖ^{−1}(S) ∪ {0} is an affine variety with an unique possible singularity at the origin 0 ∈ ℂ^{n}. Moreover S̃ is invariant under the ℂ^{∗}-action

\begin{matrix}ℂ^{∗} × S̃ & ⟶ & S̃. \\ (t, (z_{1}, …, z_{n})) & ↦ & (t^{m_{1}}z_{1}, …, t^{m_{n}}z_{n}) \end{matrix}

Let I(S̃) be the ideal of S̃,

I(S̃) = {f = f(z_{1}, …, z_{n}) ∈ ℂ[z_{1}, …, z_{n}] : f|S̃ = 0}.

Since S̃ is ℂ^{∗}-invariant, I(S̃) is generated by finitely many weighted homogeneous polynomials of weight m = (m_{1}, …, m_{n}). We shall make the second assumption.

Assumption 2. There are weighted homogeneous polynomials f_{1}, …, f_{n−2} of weight m = (m_{1}, …, m_{n}), deg f_{j} = N_{j}, such that

rank (\begin{matrix}\frac{∂f_{1}}{∂z_{1}} (z) & ⋯ & \frac{∂f_{1}}{∂z_{n}} (z) \\ ⋮ & ... & ⋮ \\ \frac{∂f_{n−2}}{∂z_{1}} (z) & ⋯ & \frac{∂f_{n−2}}{∂z_{n}} (z) \end{matrix}) = n − 2

holds for a general point z ∈ S̃.

We consider the following (1, 1)-form on W,

(1)  ω = \sqrt{−1} ∑_{i=1}^{n} \frac{1}{a_{i}^{2}} e^{−2a_{i}t}dz_{i} ∧ dz̄_{i},

where a_{i} ∈ R is a constant given by

|α|^{m_{i}} = e^{a_{i}}

and t = t(z_{1}, …, z_{n}) is a real-valued C^{∞} function defined by the equation

∑_{i=1}^{n} e^{−2a_{i}t}|z_{i}|^{2} = 1.

Since the left-hand side of the equation above is a strictly decreasing function of t for any fixed (z_{1}, …, z_{n}) ∈ W, t = t(z_{1}, …, z_{n}) is a well-defined C^{∞} function of (z_{1}, …, z_{n}). It is easy to check that ω defines a positive definite (1, 1)-form on H_{(α,m)}^{n}.

Theorem 2.1. Suppose that S satisfies Assumptions 1 and 2. Then the restriction of ω to S gives a Gauduchon metric form on S, i.e. we have:

dd^{c}ω|_{S} = 0.

Remark 2.2. If S̃ is a complete intersection in ℂ^{n}, then Assumption 2 is satisfied.

Remark 2.3. If m_{1} = ⋯ = m_{n} = 1, then the form (1) gives a Gauduchon metric on H_{(α,(1,...,1))}^{n} (see, Example 4.6) below). In general, however, (1) gives on H_{(α,m)}^{n} neither a Gauduchon metric nor a pluri-closed form, that is, dd^{c}ω does not necessarily vanish if n ≥ 3.

3 The Proof of Theorem 2.1

In this section, we prove Theorem 2.1. The following is well-known. Lemma 3.1. (Euler’s formula) Let f(z_{1}, z_{2}, …, z_{n}) be a weighted homogeneous polynomial of weight m = (m_{1}, …, m_{n}) with deg f = N. Then,

∑_{i=1}^{n} m_{i}z_{i} \frac{∂f}{∂z_{i}} (z_{1}, z_{2}, …, z_{n}) = Nf(z_{1}, z_{2}, …, z_{n}).

By a direct calculation (see Appendix A for the detail), we have (2)

∂ ¯∂ω = \sqrt{−1} ∑_{i_{1}>i_{2}>i_{3}>i_{4}} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}+a_{i_{4}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}a_{i_{4}}M^{3}} A_{i_{1}i_{2}i_{3}i_{4}}+\sqrt{−1} ∑_{i_{1},i_{2},i_{3}(i_{3}>i_{1})} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}M^{3}} B_{i_{1}i_{2}i_{3}}

where

A_{i_{1}i_{2}i_{3}i_{4}} = a_{i_{4}}^{2}|z_{i_{4}}|^{2}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})(a_{i_{1}}z_{i_{1}}dz_{i_{3}} ∧ dz_{i_{2}} + a_{i_{2}}z_{i_{2}}dz_{i_{1}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{2}} ∧ dz_{i_{1}})
×(a_{i_{1}} z̄_{i_{1}}dz̄_{i_{3}} ∧ dz̄_{i_{2}} + a_{i_{2}} z̄_{i_{2}}dz̄_{i_{1}} ∧ dz̄_{i_{3}} + a_{i_{3}} z̄_{i_{3}}dz̄_{i_{2}} ∧ dz̄_{i_{1}})
+ a_{i_{1}}^{2}|z_{i_{1}}|^{2}(a_{i_{1}} − a_{i_{2}} − a_{i_{3}} − a_{i_{4}})(a_{i_{2}}z_{i_{2}}dz_{i_{4}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{2}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{3}} ∧ dz_{i_{2}})
×(a_{i_{2}} z̄_{i_{2}}dz̄_{i_{4}} ∧ dz̄_{i_{3}} + a_{i_{3}} z̄_{i_{3}}dz̄_{i_{2}} ∧ dz̄_{i_{4}} + a_{i_{4}} z̄_{i_{4}}dz̄_{i_{3}} ∧ dz̄_{i_{2}})
+ a_{i_{2}}^{2}|z_{i_{2}}|^{2}(a_{i_{2}} − a_{i_{1}} − a_{i_{3}} − a_{i_{4}})(a_{i_{1}}z_{i_{1}}dz_{i_{4}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{1}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{3}} ∧ dz_{i_{1}})
×(a_{i_{1}} z̄_{i_{1}}dz̄_{i_{4}} ∧ dz̄_{i_{3}} + a_{i_{3}} z̄_{i_{3}}dz̄_{i_{1}} ∧ dz̄_{i_{4}} + a_{i_{4}} z̄_{i_{4}}dz̄_{i_{3}} ∧ dz̄_{i_{1}})
+ a_{i_{3}}^{2}|z_{i_{3}}|^{2}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})(a_{i_{1}}z_{i_{1}}dz_{i_{4}} ∧ dz_{i_{2}} + a_{i_{2}}z_{i_{2}}dz_{i_{1}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{2}} ∧ dz_{i_{1}})
×(a_{i_{1}} z̄_{i_{1}}dz̄_{i_{4}} ∧ dz̄_{i_{2}} + a_{i_{2}} z̄_{i_{2}}dz̄_{i_{1}} ∧ dz̄_{i_{4}} + a_{i_{4}} z̄_{i_{4}}dz̄_{i_{3}} ∧ dz̄_{i_{2}}),
B_{i_{1}i_{2}i_{3}} = (a_{i_{3}}(a_{i_{1}} + a_{i_{2}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{2}}(a_{i_{1}} + a_{i_{3}})e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} + a_{i_{1}}(a_{i_{2}} + a_{i_{3}})e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})
×a_{i_{2}}z_{i_{2}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ (a_{i_{2}} z̄_{i_{2}}dz̄_{i_{3}} ∧ dz̄_{i_{1}} + a_{i_{1}} z̄_{i_{1}}dz̄_{i_{2}} ∧ dz̄_{i_{3}} + a_{i_{3}} z̄_{i_{3}}dz̄_{i_{1}} ∧ dz̄_{i_{2}}),

and

M = ∑_{i=1}^{n} a_{i}e^{−2a_{i}t}|z_{i}|^{2}.

Thus, it is enough to show that following. Proposition 3.2. On S̃, we have

(3)  a_{i_{1}}z_{i_{1}}dz_{i_{2}} ∧ dz_{i_{3}} + a_{i_{2}}z_{i_{2}}dz_{i_{3}} ∧ dz_{i_{1}} + a_{i_{3}}z_{i_{3}}dz_{i_{1}} ∧ dz_{i_{2}} = 0.

Proof. We consider the holomorphic map

π : W ⟶ W

defined by

(z_{1}, …, z_{n}) = π(w_{1}, …, w_{n}) = (w_{1}^{m_{1}}, …, w_{n}^{m_{n}}).

Put

F_{j}(w_{1}, …, w_{n}) := π^{∗}f_{j}(z_{1}, …, z_{n}) j = 1, 2, …, n − 2.

Then F_{j} is a homogeneous polynomial of degree N_{j}. Permuting the order of coordinates z_{1}, …, z_{n}, we can assume

(4)  rank (\begin{matrix}\frac{∂f_{1}}{∂z_{1}} (z) & ⋯ & \frac{∂f_{1}}{∂z_{n−2}} (z) \\ ⋮ & ... & ⋮ \\ \frac{∂f_{n−2}}{∂z_{1}} (z) & ⋯ & \frac{∂f_{n−2}}{∂z_{n−2}} (z) \end{matrix}) = n − 2

holds at z.

Lemma 3.3. For a general point z ∈ S̃, choose w ∈ W such that z = π(w). Then we have

rank (\begin{matrix}\frac{∂F_{1}}{∂w_{1}} (w) & ⋯ & \frac{∂F_{1}}{∂w_{n−2}} (w) \\ ⋮ & ... & ⋮ \\ \frac{∂F_{n−2}}{∂w_{1}} (w) & ⋯ & \frac{∂F_{n−2}}{∂w_{n−2}} (w) \end{matrix}) = n − 2.

Proof. Using z_{i} = w_{i}^{m_{i}}, we have

\frac{∂F_{j}}{∂w_{ℓ}} = ∑_{k=1}^{n} \frac{∂f_{j}}{∂z_{k}} \frac{∂z_{k}}{∂w_{ℓ}} = m_{ℓ}w_{ℓ}^{m_{ℓ}−1} \frac{∂f_{j}}{∂z_{ℓ}} , j = 1, …, n − 2, ℓ = 1, …, n

In the matrix form, we have

(\begin{matrix}\frac{∂F_{1}}{∂w_{1}} & ⋯ & \frac{∂F_{1}}{∂w_{n}} \\ ⋮ & ... & ⋮ \\ \frac{∂F_{n−2}}{∂w_{1}} & ⋯ & \frac{∂F_{n−2}}{∂w_{n}} \end{matrix} ) = (\begin{matrix}\frac{∂f_{1}}{∂z_{1}} & ⋯ & \frac{∂f_{1}}{∂z_{n}} \\ ⋮ & ... & ⋮ \\ \frac{∂f_{n−2}}{∂z_{1}} & ⋯ & \frac{∂f_{n−2}}{∂z_{n}} \end{matrix} ) (\begin{matrix}m_{1}w_{1}^{(m_{1}−1)} & & & & & & & 0 \\ & . & . & . \\ & & & & . & . & . \\ 0 & & & & & & & m_{n}w_{n}^{(m_{n}−1)} \end{matrix})

Since

rank (\begin{matrix}\frac{∂F_{1}}{∂w_{1}} & ⋯ & \frac{∂F_{1}}{∂w_{n}} \\ ⋮ & ... & ⋮ \\ \frac{∂F_{n−2}}{∂w_{1}} & ⋯ & \frac{∂F_{n−2}}{∂w_{n}} \end{matrix} ) = n − 2

holds at w by Assumption 2, we have the lemma.

Now we go back to the proof of Proposition3.2. It is enough to show

π^{∗} (a_{i_{1}}z_{i_{1}}dz_{i_{2}} ∧ dz_{i_{3}} + a_{i_{2}}z_{i_{2}}dz_{i_{3}} ∧ dz_{i_{1}} + a_{i_{3}}z_{i_{3}}dz_{i_{1}} ∧ dz_{i_{2}}) = 0 (5) on π^{−1}(S̃). Rewriting (5) in terms of dw_{i}, we have

π^{∗}(a_{i_{1}}z_{i_{1}}dz_{i_{2}} ∧ dz_{i_{3}} + a_{i_{2}}z_{i_{2}}dz_{i_{3}} ∧ dz_{i_{1}} + a_{i_{3}}z_{i_{3}}dz_{i_{1}} ∧ dz_{i_{2}})

= a_{i_{1}}w_{i_{1}}^{m_{i_{1}}} (\frac{∂z_{i_{2}}}{∂w_{i_{2}}} dw_{i_{2}}) ∧ (\frac{∂z_{i_{3}}}{∂w_{i_{3}}} dw_{i_{3}}) + a_{i_{2}}w_{i_{2}}^{m_{i_{2}}} (\frac{∂z_{i_{3}}}{∂w_{i_{3}}} dw_{i_{3}}) ∧ (\frac{∂z_{i_{1}}}{∂w_{i_{1}}} dw_{i_{1}})

+a_{i_{3}}w_{i_{3}}^{m_{i_{3}}} (\frac{∂z_{i_{1}}}{∂w_{i_{1}}} dw_{i_{1}}) ∧ (\frac{∂z_{i_{2}}}{∂w_{i_{2}}} dw_{i_{2}})

= a_{i_{1}}w_{i_{1}}^{m_{i_{1}}} \frac{∂z_{i_{2}}}{∂w_{i_{2}}} \frac{∂z_{i_{3}}}{∂w_{i_{3}}} dw_{i_{2}} ∧ dw_{i_{3}} + a_{i_{2}}w_{i_{2}}^{m_{i_{2}}} \frac{∂z_{i_{3}}}{∂w_{i_{3}}} \frac{∂z_{i_{1}}}{∂w_{i_{1}}} dw_{i_{3}} ∧ dw_{i_{1}}

+a_{i_{3}}w_{i_{3}}^{m_{i_{3}}} \frac{∂z_{i_{1}}}{∂w_{i_{1}}} \frac{∂z_{i_{2}}}{∂w_{i_{2}}} dw_{i_{1}} ∧ dw_{i_{2}}

= a_{i_{1}}w_{i_{1}}^{m_{i_{1}}} m_{i_{2}}m_{i_{3}}w_{i_{2}}^{(m_{i_{2}}−1)}w_{i_{3}}^{(m_{i_{3}}−1)}dw_{i_{2}} ∧ dw_{i_{3}}

+a_{i_{2}}w_{i_{2}}^{m_{i_{2}}} m_{i_{3}}m_{i_{1}}w_{i_{3}}^{(m_{i_{3}}−1)}w_{i_{1}}^{(m_{i_{1}}−1)}dw_{i_{3}} ∧ dw_{i_{1}} +a_{i_{3}}w_{i_{3}}^{m_{i_{3}}} m_{i_{1}}m_{i_{2}}w_{i_{1}}^{(m_{i_{1}}−1)}w_{i_{2}}^{(m_{i_{2}}−1)}dw_{i_{1}} ∧ dw_{i_{2}}

= a_{1} \frac{m_{i_{1}}m_{i_{2}}m_{i_{3}}}{m_{1}} w_{i_{1}}^{(m_{i_{1}}−1)}w_{i_{2}}^{(m_{i_{2}}−1)}w_{i_{3}}^{(m_{i_{3}}−1)}

× (w_{i_{1}}dw_{i_{2}} ∧ dw_{i_{3}} + w_{i_{2}}dw_{i_{3}} ∧ dw_{i_{1}} + w_{i_{3}}dw_{i_{1}} ∧ dw_{i_{2}}) .

Thus, to prove (5), it is enough to show the relation

(6)  w_{i_{1}}dw_{i_{2}} ∧ dw_{i_{3}} + w_{i_{2}}dw_{i_{3}} ∧ dw_{i_{1}} + w_{i_{3}}dw_{i_{1}} ∧ dw_{i_{2}} = 0

holds at p.

Take a general point p = (p_{1}, …, p_{n}) ∈ π^{−1}(S̃) with

∏_{i=1}^{n} p_{i} ≠ 0. (7)

We introduce a new system of coordinates (u_{1}, …, u_{n}) on W by the following linear transformation,

u_{1} = \frac{∂F_{1}}{∂w_{1}} (p)w_{1} + \frac{∂F_{1}}{∂w_{2}} (p)w_{2} + ⋯ + \frac{∂F_{1}}{∂w_{n}} (p)w_{n}
u_{2} = \frac{∂F_{2}}{∂w_{1}} (p)w_{1} + \frac{∂F_{2}}{∂w_{2}} (p)w_{2} + ⋯ + \frac{∂F_{2}}{∂w_{n}} (p)w_{n}
⋮
u_{n−2} = \frac{∂F_{n−2}}{∂w_{1}} (p)w_{1} + \frac{∂F_{n−2}}{∂w_{2}} (p)w_{2} + ⋯ + \frac{∂F_{n−2}}{∂w_{n}} (p)w_{n}
u_{n−1} = w_{n−1}
u_{n} = w_{n}.

With respect to the new coordinates (u_{1}, …, u_{n}), the point p = (p_{1}, …, p_{n}) is expressed by

ṕ = u(p) = (0, …, 0, ṕ_{n−1}, ṕ_{n})

for some ṕ_{n−1} and ṕ_{n} by Euler’s Theorem. Hence we have

p_{i} = \frac{∂w_{i}}{∂u_{n−1}} (p)ṕ_{n−1} + \frac{∂w_{i}}{∂u_{n}} (p)ṕ_{n} (i = 1, 2, …, n − 2)
p_{n−1} = ṕ_{n−1}
p_{n} = ṕ_{n},

that is,

(8)  p_{i} = \frac{∂w_{i}}{∂u_{n−1}} (ṕ)ṕ_{n−1} + \frac{∂w_{i}}{∂u_{n}} (ṕ)ṕ_{n} i = 1, …, n.

Set G_{j}(u_{1}, …, u_{n}) = F_{j}(w_{1}, …, w_{n}) (j = 1, 2, …, n − 2). Note that G_{j} is also a homogeneous polynomial of degree N_{j}. Differentiating F_{j}, j = 1, …, n − 2, we have

\frac{∂F_{j}}{∂w_{1}} = ∑_{k=1}^{n} \frac{∂G_{j}}{∂u_{k}} \frac{∂u_{k}}{∂w_{1}}  = ∑_{k=1}^{n−2} \frac{∂G_{j}}{∂u_{k}} \frac{∂F_{k}}{∂w_{1}} (p)
⋮
\frac{∂F_{j}}{∂w_{n−2}} = ∑_{k=1}^{n} \frac{∂G_{j}}{∂u_{k}} \frac{∂u_{k}}{∂w_{n−2}} = ∑_{k=1}^{n−2} \frac{∂G_{j}}{∂u_{k}} \frac{∂F_{k}}{∂w_{n−2}} (p)

\frac{∂F_{j}}{∂w_{n−1}} = ∑_{k=1}^{n} \frac{∂G_{j}}{∂u_{k}} \frac{∂u_{k}}{∂w_{n−1}} = ∑_{k=1}^{n−2} \frac{∂G_{j}}{∂u_{k}} \frac{∂F_{k}}{∂w_{n−1}} (p) + \frac{∂g_{j}}{∂u_{n−1}}

\frac{∂F_{j}}{∂w_{n}} = ∑_{k=1}^{n} \frac{∂G_{j}}{∂u_{k}} \frac{∂u_{k}}{∂w_{n}}  = ∑_{k=1}^{n−2} \frac{∂G_{j}}{∂u_{k}} \frac{∂F_{k}}{∂w_{n}} (p) + \frac{∂G_{j}}{∂u_{n}} .

From this, it follows at w = p ( or u = ṕ ) that

\frac{∂G_{j}}{∂u_{k}} (ṕ) = δ_{k}^{j} (Kronecker’s delta)

for j, k = 1, …, n − 2, and

\frac{∂G_{j}}{∂u_{n−1}} (ṕ) = 0, \frac{∂G_{j}}{∂u_{n}} (ṕ) = 0

for j = 1, …, n − 2. Since F_{j} = 0, j = 1, …, n − 2, on π^{−1}(S̃), we have

dF_{j} = ∑_{k=1}^{n} \frac{∂G_{j}}{∂u_{k}} du_{k} = \frac{∂G_{j}}{∂u_{j}} du_{j} = 0.

Hence, at ṕ, we have

du_{j} = 0 j = 1, …, n − 2.

Therefore

(9)  \{ \begin{matrix}dw_{i} & = & \frac{∂w_{i}}{∂u_{n−1}} (ṕ)du_{n−1} + \frac{∂w_{i}}{∂u_{n}} (ṕ)du_{n}, i = 1, …, n − 2 \\ dw_{n−1} & = & du_{n−1} \\ dw_{n} & = & du_{n}. \end{matrix}

Put

A_{i} = \frac{∂w_{i}}{∂u_{n−1}} (ṕ), B_{i} = \frac{∂w_{i}}{∂u_{n}} (ṕ).

Then combining (8) and (9), we have

(10)  \{ \begin{matrix}dw_{i} & = & A_{i}du_{n−1} + B_{i}du_{n} \\ p_{i} & = & A_{i}ṕ_{n−1} + B_{i}ṕ_{n} \end{matrix}

for i = 1, …, n. Using the relation (10), the equality (6) can be verified at p as follows.

w_{i_{1}}dw_{i_{2}} ∧ dw_{i_{3}} + w_{i_{2}}dw_{i_{3}} ∧ dw_{i_{1}} + w_{i_{3}}dw_{i_{1}} ∧ dw_{i_{2}}
= (A_{i_{1}} ṕ_{n−1} + B_{i_{1}} ṕ_{n}) (A_{i_{2}}du_{n−1} + B_{i_{2}}du_{n}) ∧ (A_{i_{3}}du_{n−1} + B_{i_{3}}du_{n})
+ (A_{i_{2}} ṕ_{n−1} + B_{i_{2}} ṕ_{n}) (A_{i_{3}}du_{n−1} + B_{i_{3}}du_{n}) ∧ (A_{i_{1}}du_{n−1} + B_{i_{1}}du_{n})
+ (A_{i_{3}} ṕ_{n−1} + B_{i_{3}} ṕ_{n}) (A_{i_{1}}du_{n−1} + B_{i_{1}}du_{n}) ∧ (A_{i_{2}}du_{n−1} + B_{i_{2}}du_{n})
= det(Aṕ_{n−1} + Bṕ_{n}, A, B)du_{n−1} ∧ du_{n}
= 0,

where A = (A_{i_{1}}, A_{i_{2}}, A_{i_{3}}) and B = (B_{i_{1}}, B_{i_{2}}, B_{i_{3}}). Thus Proposition 3.2 is proved.

Since dd^{c} = 2\sqrt{−1}∂ ¯∂, we have by Proposition 3.2 that

(11)  π^{∗}(dd^{c}ω)|_{π^{−1}(S)} = 0.

at a general point of π^{−1}(S̃). Hence the equality (11) holds everywhere by continuity. Therefore we have

dd^{c}ω|_{S} = 0.

Thus the theorem is proved.

4 Primary Hopf surfaces

In this section, we shall give Gauduchon metrics on primary Hopf surfases which are compact quotients of ℂ^{2} \ {0} by linear infinite cyclic action groups. Here we include non-elliptic cases.

4.1 Diagonal matrix case

In this subsection, we consider a primary Hopf surface of the following form:

H_{(α,β)}^{2} = W^{2}/⟨g⟩,

where

W^{2} = C^{2} \ {(0, 0)},

g = ( \begin{matrix}α & 0 \\ 0 & β \end{matrix} ) , 0 < |α| ≤ |β| < 1.

Note that H_{(α,β)}^{2} is elliptic if and only if β^{m} = α^{n} for some m ≥ 1, n ≥ 1. Define a, b ∈ R by

|α| = e^{a}, |β| = e^{b}.

Theorem 4.1. Let t = t(z, w) be a differentiable function defined by

(12)  e^{−2at}|z|^{2} + e^{−2bt}|w|^{2} = 1.

Then the form given by

ω = \sqrt{−1}(\frac{1}{a^{2}} e^{−2at}dz ∧ d\overline{z} + \frac{1}{b^{2}} e^{−2bt}dw ∧ d\overline{w})

defines a Gauduchon metric form on H_{(α,β)}^{2}.

This theorem applies to any pair (α, β) provided that 0 < |α|, |β| < 1. Therefore, Theorem 4.1 is not included in Theorem 2.1.

Proof. Note that φ(s) = e^{−2as}|z|^{2} + e^{−2bs}|w|^{2} is an strictly increasing C^{∞}-function of s for any fixed (z, w) ≠ (0, 0). Therefore the function t = t(z, w) defined by (12) is of C^{∞} by the implicit function theorem. It is easy to check that ω is defined on H_{(α,β)}^{2} and is positive definite. Thus to prove that ω is a Gauduchon metric form, it is enough to confirm the equality

(13)  \frac{∂^{2}}{∂w∂\overline{w}} (\frac{1}{a^{2}} e^{−2at}) + \frac{∂^{2}}{∂z∂\overline{z}} (\frac{1}{b^{2}} e^{−2bt}) = 0.

First, applying \frac{∂t}{∂z}, \frac{∂t}{∂\overline{z}},\frac{∂t}{∂w}, \frac{∂t}{∂\overline{w}} to (12), we have

(14)  \frac{∂t}{∂z} = \frac{e^{−2at}\overline{z}}{M} , \frac{∂t}{∂\overline{z}} = \frac{e^{−2at}z}{M} , \frac{∂t}{∂w} = \frac{e^{−2bt}\overline{w}}{M} , \frac{∂t}{∂\overline{w}} = \frac{e^{−2bt}w}{M} ,

where

(15)  M = 2ae^{−2at}|z|^{2} + 2be^{−2bt}|w|^{2}.

Put

A = \frac{∂^{2}}{∂w∂\overline{w}} (\frac{1}{a^{2}} e^{−2at}) , B = \frac{∂^{2}}{∂z∂\overline{z}} (\frac{1}{b^{2}} e^{−2bt})

First we shall calculte A.

A = \frac{∂}{∂w} ((−\frac{2}{a}e^{−2at}) × \frac{∂t}{∂\overline{w}} )

= \frac{∂}{∂t} (−\frac{2}{a}e^{−2at}) × \frac{∂t}{∂w} × \frac{∂t}{∂\overline{w}} + (−\frac{2}{a}e^{−2at}) \frac{∂^{2}t}{∂w∂\overline{w}}

= 4e^{−2at} × \frac{e^{−4bt}|w|^{2}}{M^{2}} + (−\frac{2}{a}e^{−2at}) × \frac{∂}{∂w} (\frac{e^{−2bt}w}{M} ) .

For the 2nd term of the right-hand side of the last equation, we have

\frac{∂}{∂w} (\frac{e^{−2bt}w}{M} ) = \frac{(−2be^{−2bt} \frac{∂t}{∂w} × w + e^{−2bt})}{M}

−\frac{((−4a^{2}e^{−2at}|z|^{2} − 4b^{2}e^{−2bt}|w|^{2}) \frac{∂t}{∂w} + 2be^{−2bt}\overline{w}) × e^{−2bt}w}{M^{2}}

= \frac{e^{−2bt} − 2be^{−2bt}w \frac{∂t}{∂w}}{M} + \frac{e^{−2bt}w ((4a^{2}e^{−2at}|z|^{2} + 4b^{2}e^{−2bt}|w|^{2}) \frac{∂t}{∂w} − 2be^{−2bt}\overline{w})}{M^{2}}

= \frac{1}{M} (e^{−2bt} − 2be^{−2bt}w × \frac{e^{−2bt}\overline{w}}{M} )

+\frac{e^{−2bt}w}{M^{2}} × ((4a^{2}e^{−2at}|z|^{2} + 4b^{2}e^{−2bt}|w|^{2}) \frac{e^{−2bt}\overline{w}}{M} − 2be^{−2bt}\overline{w})

= \frac{e^{−2bt}}{M} − \frac{1}{M^{2}} × (2be^{−4bt}|w|^{2} + 2be^{−4bt}|w|^{2}) + \frac{e^{−4bt}|w|^{2}}{M^{3}} (4a^{2}e^{−2at}|z|^{2} + 4b^{2}e^{−2bt}|w|^{2}) .

Hence we obtain

A = \frac{4e^{−2(a+2b)t}|w|^{2}}{M^{2}}

−\frac{2}{a} (\frac{e^{−2(a+b)t}}{M} − \frac{e^{−2at}}{M^{2}} (4be^{−4bt}|w|^{2}) + \frac{e^{−2(a+2b)t}|w|^{2}}{M^{3}} (4a^{2}e^{−2at}|z|^{2} + 4b^{2}e^{−2bt}|w|^{2}))

= \frac{4e^{−2(a+2b)t}|w|^{2}}{M^{2}}

−\frac{2e^{−2(a+b)t}}{aM} + \frac{8be^{−2(a+2b)t}|w|^{2}}{aM^{2}} − \frac{8e^{−2(a+2b)t}|w|^{2}}{aM^{3}} (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2}) .

Next, we shall calculate B.

B = \frac{∂}{∂z} ((−\frac{2}{b}e^{−2bt}) × \frac{∂t}{∂\overline{z}} ) = \frac{∂}{∂t} (−\frac{2}{b}e^{−2bt}) × \frac{∂t}{∂z} × \frac{∂t}{∂\overline{z}} + (−\frac{2}{b}e^{−2bt}) \frac{∂^{2}t}{∂z∂\overline{z}}

= \frac{4e^{−2bt}e^{−4at}|z|^{2}}{M^{2}} − (\frac{2}{b}e^{−2bt}) × \frac{∂}{∂z} (\frac{e^{−2at}z}{M} )

For the 2nd term of the right-hand side of the last equation, we have

\frac{∂}{∂z} (\frac{e^{−2at}z}{M} ) = \frac{−2ae^{−2at}z × \frac{∂t}{∂z} × M + e^{−2at}M − (\frac{∂}{∂z} M) × e^{−2at}z}{M^{2}}

= \frac{1}{M} (−2ae^{−2at}z × \frac{∂t}{∂z} + e^{−2at})

−\frac{e^{−2at}z}{M^{2}} ((−4a^{2}e^{−2at}|z|^{2} − 4b^{2}e^{−2bt}|w|^{2}) \frac{∂t}{∂z} + (2ae^{−2at}\overline{z}))

= \frac{e^{−2at}}{M} − \frac{2ae^{−2at}z}{M} × \frac{e^{−2at}\overline{z}}{M}

+\frac{e^{−2at}z}{M^{2}} × (4a^{2}e^{−2at}|z|^{2} + 4b^{2}e^{−2bt}|w|^{2}) × \frac{e^{−2at}\overline{z}}{M} − \frac{2ae^{−4at}|z|^{2}}{M^{2}}

= \frac{e^{−2at}}{M} − \frac{4ae^{−4at}|z|^{2}}{M^{2}} + \frac{e^{−4at}|z|^{2} (4a^{2}e^{−2at}|z|^{2} + 4b^{2}e^{−2bt}|w|^{2})}{M^{3}}.

Therefore, we obtain

B = \frac{4e^{−2(2a+b)t}|z|^{2}}{M^{2}}

−\frac{2e^{−2(a+b)t}}{bM} + \frac{8ae^{−2(2a+b)t}|z|^{2}}{bM^{2}} − \frac{8e^{−2(2a+b)t}|z|^{2}}{bM^{3}} (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2}) .

Thus the left-hand side of (13) becomes

\frac{∂^{2}}{∂w∂\overline{w}} (\frac{1}{a^{2}} e^{−2at}) + \frac{∂^{2}}{∂z∂\overline{z}} (\frac{1}{b^{2}} e^{−2bt}) = A + B

= \frac{4e^{−2(a+2b)t}|w|^{2}}{M^{2}} − \frac{2e^{−2(a+b)t}}{aM} + \frac{8be^{−2(a+2b)t}|w|^{2}}{aM^{2}} − \frac{8e^{−2(a+2b)t}|w|^{2}}{aM^{3}} (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2})

+\frac{4e^{−2(2a+b)t}|z|^{2}}{M^{2}} − \frac{2e^{−2(a+b)t}}{bM} + \frac{8ae^{−2(2a+b)t}|z|^{2}}{bM^{2}} − \frac{8e^{−2(2a+b)t}|z|^{2}}{bM^{3}} (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2})

= −\frac{2e^{−2(a+b)t} (a + b)}{abM} + \frac{4be^{−2(a+2b)t}|w|^{2} (a + 2b) + 4ae^{−2(2a+b)t}|z|^{2} (2a + b)}{abM^{2}}

−\frac{8 (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2})}{abM^{3}} × (be^{−2(a+2b)t}|w|^{2} + ae^{−2(2a+b)t}|w|^{2})

= \frac{2e^{−2(a+b)t}}{abM^{3}} (− (a + b) M^{2} + (2be^{−2bt}|w|^{2} (a + 2b) + 2ae^{−2at}|z|^{2} (2a + b)) M)

−\frac{2e^{−2(a+b)t}}{abM^{3}} (4 (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2}) × (ae^{−2at}|z|^{2} + be^{−2bt}|w|^{2})) = (⋆⋆).

Here we recall the equality (15). Then,

(⋆⋆) = \frac{2e^{−2(a+b)t}}{abM^{3}} (− (a + b) M^{2} + (2be^{−2bt}|w|^{2} (a + 2b) + 2ae^{−2at}|z|^{2} (2a + b))M)

−\frac{2e^{−2(a+b)t}}{abM^{3}} (2 (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2}) M)

= \frac{2e^{−2(a+b)t}}{abM^{2}} (− (a + b) M + (2be^{−2bt}|w|^{2} (a + 2b) + 2ae^{−2at}|z|^{2} (2a + b))
−2 (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2})).

Thus we have to show the vanishing of

C := − (a + b) M + (2be^{−2bt}|w|^{2} (a + 2b) + 2ae^{−2at}|z|^{2} (2a + b)) − 2 (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2})

− (a + b) M + (2be^{−2bt}|w|^{2} (a + 2b) + 2ae^{−2at}|z|^{2} (2a + b)) − 2 (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2})

= − (a + b) (2ae^{−2at}|z|^{2} + 2be^{−2bt}|w|^{2}) + (2be^{−2bt}|w|^{2} (a + 2b) + 2ae^{−2at}|z|^{2} (2a + b))
−2 (a^{2}e^{−2at}|z|^{2} + b^{2}e^{−2bt}|w|^{2})
= − (a + b) (2ae^{−2at}|z|^{2} + 2be^{−2bt}|w|^{2}) + 2 (a + 2b) be^{−2bt}|w|^{2} + 2a (2a + b) e^{−2at}|z|^{2}
−2a^{2}e^{−2at}|z|^{2} − 2b^{2}e^{−2bt}|w|^{2}
= (−2a (a + b) + 2a (2a + b) − 2a^{2}) e^{−2at}|z|^{2} + (−2b (a + b) + 2 (a + 2b) − 2b^{2}) e^{−2bt}|w|^{2}.

Here we have

−2a (a + b) + 2a (2a + b) − 2a^{2} = −2a^{2} − 2ab + 4a^{2} + 2ab − 2a^{2} = 0. and

−2b (a + b) + 2 (a + 2b) − 2b^{2} = −2ab − 2b^{2} + 2ab + 4b^{2} − 2b^{2} = 0.

Hence C = 0. Thus we have completed the calculation and obtained the theorem.

4.2 Non-Diagonal matrix case

In this subsection, we shall consider a primary Hopf surface of the following form:

H_{(α,α,1)}^{2} = W^{2}/⟨g⟩,

where

W^{2} = C^{2} \ {(0, 0)},

g = ( \begin{matrix}α & 1 \\ 0 & α \end{matrix} ) , α ∈ C such that 0 < |α| < 1.

Note that the surface H_{(α,α,1)}^{2} is not an elliptic surface. Theorem 4.2. Let a, b be a real number such that a > 0, b > 5 and set Q = log |α|^{2}(< 0). Let t = t(z, w) be a differentiable function of (z, w) defined by

\frac{1}{b − 4}|α|^{2}Q^{2} |α^{t}z + tα^{t−1}w|^{2} + |α^{t}w|^{2} = 1. (16)

Then the form given by

(17) ω = \sqrt{−1} (a|α|^{2t}dz ∧ dz̄ + (t − \frac{2}{Q}) aᾱ^{−1}|α|^{2t}dz ∧ dw̄

+ (t − \frac{2}{Q}) aα^{−1}|α|^{2t}dw ∧ dz̄ + a (t^{2} − \frac{4}{Q}t + \frac{b}{Q^{2}} ) |α|^{2(t−1)}dw ∧ dw̄)

defines a Gauduchon metric form on H_{(α,α,1)}^{2}.

To prove Theorem 4.2, we shall prepare several lemmas.

Lemma 4.3. Set

F (t) = \frac{1}{b − 4}|α|^{2}Q^{2}|α^{t}z + tα^{t−1}w|^{2} + |α^{t}w|^{2}.

Then, for any (z, w) ∈ C^{2} \ {0}, F (t) is a strictly monotone decreasing function. Proof. If w = 0, then

F (t) = \frac{1}{b − 4}|α|^{2}Q^{2}|α^{t}z|^{2}
= \frac{1}{b − 4}|α|^{2}Q^{2}|z|^{2}|α|^{2t}.

Hence, since z ≠ 0, we have

F′(t) = \frac{1}{b − 4}|α|^{2}Q^{2}|z|^{2}|α|^{2t} log |α|^{2}
= \frac{1}{b − 4}|α|^{2}Q^{3}|z|^{2}|α|^{2t} < 0.

Therefore, F (t) is a strictly monotone decreasing function. Next, consider the case w ≠ 0. Set B = \frac{1}{b−4}|α|^{2}Q^{2} (> 0).

F (t) = B|α^{t}z + tα^{t−1}w|^{2} + |α^{t}w|^{2}
= (B|z|^{2} + |w|^{2})|α|^{2t} + tB(ᾱ^{−1}zw̄ + α^{−1}wz̄)|α|^{2t} + B|w|^{2}t^{2}|α|^{2(t−1)}.
F′(t) = (B|z|^{2} + |w|^{2})|α|^{2t}Q + B(ᾱ^{−1}zw̄ + α^{−1}wz̄)|α|^{2t} + tB(ᾱ^{−1}zw̄ + α^{−1}wz̄)|α|^{2t}Q
+2B|w|^{2}t|α|^{2(t−1)} + B|w|^{2}t^{2}|α|^{2(t−1)}Q
= |α|^{2t} (B|w|^{2}|α|^{−2}Q · t^{2} + B ((ᾱ^{−1}zw̄ + α^{−1}wz̄)Q + 2|w|^{2}|α|^{−2}) t

+ ((B|z|^{2} + |w|^{2})Q + B(ᾱ^{−1}zw̄ + α^{−1}wz̄)))

= |α|^{2t} (B|w|^{2}|α|^{−2}Q (t + \frac{1}{2} \frac{B((ᾱ^{−1}zw̄ + α^{−1}wz̄)Q + 2|w|^{2}|α|^{−2})}{B|w|^{2}|α|^{−2}Q})^{2}

− (\frac{B^{2}((ᾱ^{−1}zw̄ + α^{−1}wz̄)Q + 2|w|^{2}|α|^{−2})^{2}}{4B|w|^{2}|α|^{−2}Q} ) − ((B|z|^{2} + |w|^{2})Q + B(ᾱ^{−1}zw̄ + α^{−1}wz̄)))

= |α|^{2t} (B|w|^{2}|α|^{−2}Q (t + \frac{1}{2} \frac{(ᾱ^{−1}zw̄ + α^{−1}wz̄)Q + 2|w|^{2}|α|^{−2}}{|w|^{2}|α|^{−2}Q} )^{2}

− \frac{B((ᾱ^{−1}zw̄ + α^{−1}wz̄)Q + 2|w|^{2}|α|^{−2})^{2} − 4|w|^{2}|α|^{−2}Q ((B|z|^{2} + |w|^{2})Q + B(ᾱ^{−1}zw̄ + α^{−1}wz̄))}{4|w|^{2}|α|^{−2}Q})

= |α|^{2t} (B|w|^{2}|α|^{−2}Q (t + \frac{1}{2} \frac{(ᾱ^{−1}zw̄ + α^{−1}wz̄)Q + 2|w|^{2}|α|^{−2}}{|w|^{2}|α|^{−2}Q} )^{2}

− \frac{B(ᾱ^{−1}zw̄ − α^{−1}wz̄)^{2}Q + 4|w|^{4}|α|^{−2}Q (\frac{1}{b−4} − 1)}{4|w|^{2}|α|^{−2}} ) .

Here,

Q < 0
(ᾱ^{−1}zw̄ − α^{−1}wz̄)^{2} ≤ 0
\frac{1}{b − 4} − 1 < 0.

Therefore, F′(t) < 0, and F (t) is a strictly monotone decreasing function.

From Lemma4.3, it follows that the function t is defined as a C^{∞}-function at (z, w) by the equation (16). Lemma 4.4. The form (17) is well-defined on H_{(α,α,1)}^{2}. Proof. The equality (g^{s})^{∗}ω = ω, where

g^{s} : (z, w) ↦ (α^{s}z + sα^{s−1}w, α^{s}w), s ∈ ℝ.

Actually,

(g^{s})^{∗}ω = \sqrt{−1} · a|α|^{2t}((α^{s}dz + sα^{s−1}dw) ∧ (ᾱ^{s}dz̄ + sᾱ^{s−1}dw̄)

+(t − \frac{2}{Q})ᾱ^{−1}(α^{s}dz + sα^{s−1}dw) ∧ (ᾱ^{s}dw̄)

+(t − \frac{2}{Q})α^{−1}(α^{s}dw) ∧ (ᾱ^{s}dz̄ + sᾱ^{s−1}dw̄) + (t^{2} − \frac{4}{Q}t + \frac{b}{Q^{2}} )|α|^{−2}|α|^{2s}dw ∧ dw̄)

= \sqrt{−1}(a|α|^{2(t+s)}dz ∧ dz̄ + (a|α|^{2t} · sα^{s}ᾱ^{s−1} + (t − \frac{2}{Q})aᾱ^{−1}|α|^{2(t+s)})dz ∧ dw̄

+(a|α|^{2t} · sα^{s−1}ᾱ^{s} + (t − \frac{2}{Q})aα^{−1}|α|^{2(t+s)})dw ∧ dz̄

+(as^{2}|α|^{2(t+s−1)} + a(t − \frac{2}{Q})s|α|^{2(t+s−1)} + a(t − \frac{2}{Q})s|α|^{2(t+s−1)}

+a(t^{2} − \frac{4}{Q}t + \frac{b}{Q^{2}} )|α|^{2(t+s−1)})dw ∧ dw̄)

= \sqrt{−1}(a|α|^{2(t+s)}dz ∧ dz̄ + a|α|^{2(t+s)}(sᾱ^{−1} + (t − \frac{2}{Q})ᾱ^{−1})dz ∧ dw̄

+a|α|^{2(t+s)}(sα^{−1} + (t − \frac{2}{Q})α^{−1})dw ∧ dz̄

+a|α|^{2(t+s−1)}(s^{2} + (t − \frac{2}{Q})s + (t − \frac{2}{Q})s + (t^{2} − \frac{4}{Q}t + \frac{b}{Q^{2}} ))dw ∧ dw̄)

= \sqrt{−1}(a|α|^{2(t+s)}dz ∧ dz̄ + a((t + s) − \frac{2}{Q})ᾱ^{−1}|α|^{2(t+s)}dz ∧ dw̄

+a((t + s) − \frac{2}{Q})α^{−1}|α|^{2(t+s)}dw ∧ dz̄

+a((t + s)^{2} − \frac{2}{Q}(t + s) + \frac{b}{Q^{2}} )|α|^{2(t+s−1)}dw ∧ dw̄),

and

\frac{1}{b − 4}|α|^{2}Q^{2} · |α^{t}(α^{s}z + sα^{s−1}w) + tα^{t−1}(α^{s}w)|^{2} + |α^{t}(α^{s}w)|^{2} = 1

⟺ \frac{1}{b − 4}|α|^{2}Q^{2}|α^{t+s}z + (s + t)α^{(t+s)−1}w|^{2} + |α^{(}t + s)w|^{2} = 1.

Hence, (g^{s})^{∗}ω is substitution of (t + s) for t of ω. So, the form (17) is invariant under the action of the group ⟨g⟩.

Lemma 4.5. The form (17) defines a Harmitian metric on H_{(α,α,1)}^{2}.

Proof. It is clear that the coefficient of (17) is a Hermitian symmetric. So, it is enough to show that (17) is positive definite.

a|α|^{2t}|ζ_{1}|^{2} + a(t − \frac{2}{Q})ᾱ^{−1}|α|^{2t}ζ_{1}ζ̄_{2} + a(t − \frac{2}{Q})α^{−1}|α|^{2t}ζ_{2}ζ̄_{1}

+a(t^{2} − \frac{4}{Q}t + \frac{b}{Q^{2}} )|α|^{2(t−1)}|ζ_{2}|^{2}

= a|α|^{2t}(|ζ_{1}|^{2} + (t − \frac{2}{Q})ᾱ^{−1}ζ_{1}ζ̄_{2} + (t − \frac{2}{Q})α^{−1}ζ_{2}ζ̄_{1} + ((t^{2} − \frac{4}{Q}t + \frac{4}{Q^{2}} ) + \frac{b − 4}{Q^{2}}) |α|^{−2}|ζ_{2}|^{2})

= a|α|^{2t}(|ζ_{1}|^{2} + (t − \frac{2}{Q})ᾱ^{−1}ζ_{1}ζ̄_{2} + (t − \frac{2}{Q})α^{−1}ζ_{2}ζ̄_{1} + (t − \frac{2}{Q})^{2}|α|^{−2}|ζ_{2}|^{2} + \frac{b − 4}{Q^{2}}|α|^{−2}|ζ_{2}|^{2})

= a|α|^{2t}((ζ_{1} + (t − \frac{2}{Q}) α^{−1}ζ_{2}) (ζ̄_{1} + (t − \frac{2}{Q}) ᾱ^{−1}ζ̄_{2}) + \frac{b − 4}{Q^{2}}|α|^{−2}|ζ_{2}|^{2})

= a|α|^{2t}(|ζ_{1} + (t − \frac{2}{Q}) α^{−1}ζ_{2}|^{2} + \frac{b − 4}{Q^{2}}|α|^{−2}|ζ_{2}|^{2}) ≥ 0

and equality if and only if ζ = (ζ_{1}, ζ_{2}) = (0, 0). Hence, (17) defines a Harmitian metric on H_{(α,α,1)}^{2}.

Proof of Theorem 4.2. The required equality dd^{c}ω = 0 can be verified by a direct rather long calculation similar to the proof of Theorem 4.1, which will be carried out in Appendix B.

Example 4.6. Recall Remark 2.3. Fix any α ∈ C such that 0 < |α| < 1. Let

H_{(α,(1,1,...,1))}^{n} = W^{n}/⟨g⟩,

where

W^{n} = C^{n} \ {0},

and g is defined by

g(z_{1}, …, z_{n}) = (αz_{1}, …, αz_{n}).

Then the hermitian metric form

(18)  ω = \sqrt{−1}\frac{∑_{i=1}^{n} dz_{i} ∧ dz̄_{i}}{∑_{i=1}^{n} |z_{i}|^{2}} .

gives a Gauduchon metric form on H_{(α,(1,1,...,1))}^{n}. In deed, the equality dd^{c}ω^{n−1} = 0

holds. To verify this, we set M = ∑_{i=1}^{n} |z_{i}|^{2}. Then,

ω^{n−1} = (\sqrt{−1}\frac{∑_{i_{1}=1}^{n} dz_{i_{1}} ∧ dz̄_{i_{1}}}{M} ) ∧ ⋯ ∧ (\sqrt{−1}\frac{∑_{i_{n−1}=1}^{n} dz_{i_{n−1}} ∧ dz̄_{i_{n−1}}}{M} )

= (\sqrt{−1})^{n−1} \frac{(n − 1)!}{M^{n−1}} ∑_{i=1}^{n}(dz_{1} ∧ dz̄_{1}) ∧ ⋯ ∧ ˆ (dz_{i} ∧ d̂z̄_{i}) ∧ ⋯ ∧ (dz_{n} ∧ dz̄_{n}),

where dz_{1} ∧ ⋯ ∧ d̂z_{i} ∧ ⋯ ∧ dz_{n} = dz_{1} ∧ ⋯ ∧ dz_{i−1} ∧ dz_{i+1} ∧ ⋯ ∧ dz_{n}. Hence,

¯∂ω^{n−1} = (\sqrt{−1})^{n−1}(n − 1)! ∑_{i=1}^{n} \frac{−(n − 1)M^{n−2}z_{i}}{M^{2(n−1)}} dz̄_{i} ∧ (dz_{1} ∧ dz̄_{1}) ∧ ⋯

⋯ ∧ ˆ (dz_{i} ∧ d̂z̄_{i}) ∧ ⋯ ∧ (dz_{n} ∧ dz̄_{n})

= −(n − 1)(\sqrt{−1})^{n−1}(n − 1)! ∑_{i=1}^{n} \frac{z_{i}}{M^{n}} (dz_{1} ∧ dz̄_{1}) ∧ ⋯

⋯ ∧ (d̂z_{i} ∧ dz̄_{i}) ∧ ⋯ ∧ (dz_{n} ∧ dz̄_{n}).

∂ ¯∂ω^{n−1} = −(n − 1)(\sqrt{−1})^{n−1}(n − 1)! ∑_{i=1}^{n} \frac{M^{n} − n · M^{n−1}z_{i} · z̄_{i}}{M^{2n}} dz_{i} ∧ (dz_{1} ∧ dz̄_{1}) ∧ ⋯

⋯ ∧ ˆ (dz_{i} ∧ dz̄_{i}) ∧ ⋯ ∧ (dz_{n} ∧ dz̄_{n})

= −(n − 1)(\sqrt{−1})^{n−1}(n − 1)! ∑_{i=1}^{n} \frac{M − n · |z_{i}|^{2}}{M^{n+1}} (dz_{1} ∧ dz̄_{1}) ∧ ⋯ ∧ (dz_{n} ∧ dz̄_{n})

= \frac{−(n − 1)(\sqrt{−1})^{n−1}(n − 1)!}{M^{n+1}} (∑_{i=1}^{n} M − n ∑_{i=1}^{n} |z_{i}|^{2}) (dz_{1} ∧ dz̄_{1}) ∧ ⋯ ∧ (dz_{n} ∧ dz̄_{n})

= \frac{−(n − 1)(\sqrt{−1})^{n−1}(n − 1)!}{M^{n+1}} (n · M − n · M) (dz_{1} ∧ dz̄_{1}) ∧ ⋯ ∧ (dz_{n} ∧ dz̄_{n})
= 0.

Therefore, (18) gives a Gauduchon metric form on H_{(α,(1,1,...,1))}^{n}.

5 Inoue surfaces

In [I]. Inoue introduced three kinds of VII_{o} surfaces with vanishing second Betti numbers. We recall the definitions of these surfaces and construct Gauduchon metric forms on each surface. By H we denote the upper half space of the complex plane.

5.1 Surface S_{M}

Definition 5.1. Let M = (m_{ij}) ∈ SL(3, Z) be a unimodular matrix with three distinct eigenvalues α > 1, β ≠ β̄ ∈ C. Let (a_{1}, a_{2}, a_{3}) ∈ R^{3}, (b_{1}, b_{2}, b_{3}) ∈ C^{3} be eigenvectors corresponding to α and β, respectively. Let G_{M} be the group of analytic automorphisms of H × C generated by:

g_{0} : (w, z) ⟼ (αw, βz)
g_{i} : (w, z) ⟼ (w + a_{i}, z + b_{i})  (i = 1, 2, 3).

The action of G_{M} on H × C is properly discontinuous and fixed point free and S_{M} := H × C/ G_{M} becomes a compact complex surface.

Definition 5.2. Let N = (n_{ij}) ∈ SL(2, Z) be a unimodular matrix with two distinct eigenvalues α > 1/α, and let (a_{1}, a_{2}) and (b_{1}, b_{2}) be real eigenvectors corresponding to α and 1/α, respectively. We fix integers p, q, r (r ≠ 0) and a complex number t, and define c_{1}, c_{2} ∈ R to be the solution of the following equation:

(c_{1}, c_{2}) = (c_{1}, c_{2}) ·^{t}N + (e_{1}, e_{2}) + \frac{b_{1}a_{2} − b_{2}a_{1}}{r} (p, q),

where e_{i} := \frac{1}{2}n_{i1}(n_{i1} − 1)a_{1}b_{1} + \frac{1}{2}n_{i2}(n_{i2} − 1)a_{2}b_{2} + n_{i1}n_{i2}b_{1}a_{2} (i = 1, 2). Let G_{N,p,q,r;t}^{(+)} be the group of analytic automorphisms of H × C generated by:

g_{0} : (w, z) ⟼ (αw, z + t)
g_{i} : (w, z) ⟼ (w + a_{i}, z + b_{i}w + c_{i})  (i = 1, 2)
g_{3} : (w, z) ⟼ (w, z + \frac{b_{1}a_{2} − b_{2}a_{1}}{r} ).

The action of G_{N,p,q,r;t}^{(+)} on H × C is properly discontinuous and fixed point free. We

define S_{N,p,q,r;t}^{(+)} to be the quotient surface H × C/G_{N,p,q,r;t}^{(+)}.

Definition 5.3. Let N = (n_{ij}) ∈ GL(2, Z) be a 2×2 matrix with det(N) = −1 and two distinct real eigenvalues α > 1 > −1/α, and let (a_{1}, a_{2}) and (b_{1}, b_{2}) be real eigenvectors corresponding to α and −1/α, respectively. We fix integers p, q, r (r ≠ 0). We define c_{1}, c_{2} ∈ R to be the solution of the following equation:

−(c_{1}, c_{2}) = (c_{1}, c_{2}) ·^{t}N + (e_{1}, e_{2}) + \frac{b_{1}a_{2} − b_{2}a_{1}}{r} (p, q),

where e_{i} := \frac{1}{2}n_{i1}(n_{i1} − 1)a_{1}b_{1} + \frac{1}{2}n_{i2}(n_{i2} − 1)a_{2}b_{2} + n_{i1}n_{i2}b_{1}a_{2} (i = 1, 2). Let G_{N,p,q,r}^{(−)} be the group of analytic automorphisms of H × C generated by:

g_{0} : (w, z) ⟼ (αw, −z)
g_{i} : (w, z) ⟼ (w + a_{i}, z + b_{i}w + c_{i})  (i = 1, 2)
g_{3} : (w, z) ⟼ (w, z + \frac{b_{1}a_{2} − b_{2}a_{1}}{r} ).

The action of G_{N,p,q,r}^{(−)} on H × C is properly discontinuous and fixed point free. We

define a surface S_{N,p,q,r}^{(−)} to be the quotient surface H × C/G_{N,p,q,r}^{(−)}.

We consider the Hermitian metric form on H × C defined by:

(19)  ω = \sqrt{−1} (\frac{dw ∧ dw̄}{(Im(w))^{2}} + Im(w)dz ∧ dz̄) ,

where Im(w)>0. Then, we prove:

Lemma 5.4. The Hermitian metric form ω given by (19) is invariant under the action of the group G_{M}. Proof. Note that a_{1}, a_{2}, α are real, and α · β · β̄ = detM = 1. Therefore it is clear that g_{j}^{∗}ω = ω holds for j = 1, 2, 3. For g_{0}, we have

g_{0}^{∗}ω = \sqrt{−1}(\frac{|α|^{2}}{(Im(αw))^{2}} dw ∧ dw̄ + (Im(αw))|β|^{2}dz ∧ dz̄)
= \sqrt{−1}(\frac{α^{2}}{α^{2}(Im(w))^{2}} dw ∧ dw̄ + α|β|^{2}(Im(w))dz ∧ dz̄)
= ω.

Hence, (19) is a well-defined Hermitian metric form on S_{M}.

Furthermore, we have the following Theorem 5.5. The form ω given by (19) defines a Gauduchon metric form on S_{M}. Proof. This is claer by

dd^{c}ω = 2\sqrt{−1}∂ ¯∂ω = −2∂ ¯∂(\frac{dw ∧ dw̄}{(Im(w))^{2}} + Im(w)dz ∧ dz̄) = −2∂ ¯∂(Im(w)dz ∧ dz̄)
= −\sqrt{−1}∂(dw̄ ∧ dz ∧ dz̄) = 0.

5.2 Surfaces S_{N,p,q,r;t}^{(+)} and S_{N,p,q,r;t}^{(−)}

We define the Hermitian metric form on H × C given by:

(20) ω = \sqrt{−1} (\frac{1 + (Im(z))^{2}}{(Im(w))^{2}} dw ∧ dw̄ − \frac{Im(z)}{Im(w)} (dw ∧ dz̄ + dz ∧ dw̄) + dz ∧ dz̄) ,

where Im(w)>0. In fact, the coefficient of (20) is Hermitian symmetric and

\frac{1 + (Im(z))^{2}}{(Im(w))^{2}} |ζ_{2}|^{2} − \frac{Im(z)}{Im(w)} (ζ_{2}ζ̄_{1} + ζ_{1}ζ̄_{2}) + |ζ_{1}|^{2}

= \frac{1}{(Im(w))^{2}} ((1 + (Im(z))^{2}) |ζ_{2}|^{2} − Im(z)Im(w) (ζ_{2}ζ̄_{1} + ζ_{1}ζ̄_{2}) + (Im(w))^{2}|ζ_{1}|^{2})
= \frac{1}{(Im(w))^{2}} (|ζ_{2}|^{2} + |Im(z)ζ_{2} − Im(w)ζ_{1}|^{2}) ≥ 0,

i.e. (20) is positive definite. Therefore (20) is Hermitian metric form on H × C. Lemma 5.6. The Hermitian metric form ω on H×C defined by (20) is invariant under the action of G_{N,p,q,r}^{(−)}. Similarly, it is invariant under the action of G_{N,p,q,r;t}^{(+)}, if t ∈ ℝ. Proof. First, we shall prove that (20) is G_{N,p,q,r}^{(−)}-invariant.

g_{0}^{∗}ω = \sqrt{−1} (\frac{1 + (Im(−z))^{2}}{(Im(αw))^{2}} |α|^{2}dw ∧ dw̄ + \frac{Im(−z)}{Im(αw)} (αdw ∧ dz̄ + αdz ∧ dw̄) + dz ∧ dz̄)
= \sqrt{−1} (\frac{1 + (Im(z))^{2}}{α^{2}(Im(w))^{2}} α^{2}dw ∧ dw̄ − \frac{Im(z)}{αIm(w)}α (dw ∧ dz̄ + dz ∧ dw̄) + dz ∧ dz̄)
= ω,

g_{j}^{∗}ω = \sqrt{−1} (\frac{1 + (Im(z + b_{i}w + c_{i}))^{2}}{(Im(w + a_{i}))^{2}} dw ∧ dw̄ − \frac{Im(z + b_{i}w + c_{i})}{Im(w + a_{i})}(dw ∧ d\overline{(z + b_{i}w + c_{i})}

+ d(z + b_{i}w + c_{i}) ∧ dw̄) + d(z + b_{i}w + c_{i}) ∧ d\overline{(z + b_{i}w + c_{i})})

= \sqrt{−1} (\frac{1 + (Im(z))^{2} + 2b_{i}Im(z)Im(w) + b_{i}^{2}(Im(w))^{2}}{(Im(w))^{2}} dw ∧ dw̄

− \frac{Im(z) + b_{i}Im(w)}{Im(w)} (dw ∧ (dz̄ + b_{i}dw̄) + (dz + b_{i}dw) ∧ dw̄) + (dz + b_{i}dw) ∧ (dz̄ + b_{i}dw̄))

= \sqrt{−1} (\frac{1 + (Im(z))^{2}}{(Im(w))^{2}} dw ∧ dw̄ − \frac{Im(z)}{Im(w)} (dw ∧ dz̄ + dz ∧ dw̄) + dz ∧ dz̄)
= ω  (j = 1, 2),

g_{3}^{∗}ω = \sqrt{−1} (\frac{1 + (Im(z + \frac{b_{1}a_{2}−b_{2}a_{1}}{r}))^{2}}{(Im(w))^{2}} dw ∧ dw̄

− \frac{Im(z + \frac{b_{1}a_{2}−b_{2}a_{1}}{r})}{Im(w)} (dw ∧ dz̄ + dz ∧ dw̄) + dz ∧ dz̄)

= \sqrt{−1} (\frac{1 + (Im(z))^{2}}{(Im(w))^{2}} dw ∧ dw̄ − \frac{Im(z)}{Im(w)} (dw ∧ dz̄ + dz ∧ dw̄) + dz ∧ dz̄)
= ω.

Thus (20) is G_{N,p,q,r}^{(−)}-invariant. Similarly, we can check that (20) is G_{N,p,q,r;t}^{(+)}-invariant under the assumption that t is a real number.

Thus (20) defines a Hermitian metric form on S_{N,p,q,r}^{(−)}, and also on S_{N,p,q,r;t}^{(+)} provided that t is real.

Theorem 5.7. The form ω given by (20) defines a Gauduchon metric form on S_{N,p,q,r}^{(−)}.

Similarly, it defines a Gauduchon metric form on S_{N,p,q,r;t}^{(+)}, if t is real.

Proof. This is clear by the following calculation.

dd^{c}ω = 2\sqrt{−1}∂ ¯∂ω

= −2∂ ¯∂(\frac{1 + (Im(z))^{2}}{(Im(w))^{2}} dw ∧ dw̄ − \frac{Im(z)}{Im(w)} (dw ∧ dz̄ + dz ∧ dw̄) + dz ∧ dz̄)
= −2∂ (\frac{1}{(Im(w))^{2}} (2Im(z) · \frac{\sqrt{−1}}{2}) dz̄ ∧ dw ∧ dw̄ − \frac{Im(z)}{Im(w)} · \frac{\sqrt{−1}}{2}dz̄ ∧ dz ∧ dw̄

+ \frac{Im(z)}{(Im(w))^{2}} · \frac{i}{2}dw̄ ∧ dw ∧ dz̄)

= −2∂ (\sqrt{−1}\frac{Im(z)}{(Im(w))^{2}} dz̄ ∧ dw ∧ dw̄ − \frac{\sqrt{−1}}{2} \frac{Im(z)}{Im(w)}dz̄ ∧ dz ∧ dw̄

+ \frac{\sqrt{−1}}{2} \frac{Im(z)}{(Im(w))^{2}} dw̄ ∧ dw ∧ dz̄)

= −2 (\frac{1}{2} \frac{1}{(Im(w))^{2}} dz ∧ dz̄ ∧ dw ∧ dw̄ + \frac{1}{4} \frac{1}{(Im(w))^{2}} dw ∧ dz̄ ∧ dz ∧ dw̄

+ \frac{1}{4} \frac{1}{(Im(w))^{2}} dz ∧ dw̄ ∧ dw ∧ dz̄)

= −2 (\frac{1}{2} \frac{1}{(Im(w))^{2}} − \frac{1}{4} \frac{1}{(Im(w))^{2}} − \frac{1}{4} \frac{1}{(Im(w))^{2}} ) dz ∧ dz̄ ∧ dw ∧ dw̄
= 0 .

Remark 5.8. The forms (19) and (20) were first given by Triccerri in [Tr] as explicit examples of locally conformal Kähler metrics on Inoue surfaces. We found that these forms actually give Gauduchon metrics on these surfaces by simple calculations. We don’t know, however, that, for non-real t, how to construct Gauduchon metric forms on S_{N,p,q,r;t}^{(+)}.

Appendix A

We carry out the calculation to derive (2) appeared in section 3. We replace here ω with \sqrt{−1}ω to avoid useless complexity of notation.

Set M = ∑_{i=1}^{n} a_{i}e^{−2a_{i}t}|z_{i}|^{2}, and N = ∑_{i=1}^{n} a_{i}^{2}e^{−2a_{i}t}|z_{i}|^{2}.

\overline{∂}ω = ∑_{i_{1}=1}^{n} \frac{1}{a_{i_{1}}^{2}} (−2a_{i_{1}}e^{−2a_{i_{1}}t} ∑_{i_{2}=1}^{n} \frac{∂t}{∂\overline{z}_{i_{2}}} )d\overline{z}_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{1}}
∂\overline{∂}ω = ∑_{i_{1}=1}^{n}(−\frac{2}{a_{i_{1}}} )((−2a_{i_{1}}e^{−2a_{i_{1}}t}) ∑_{i_{3}=1}^{n} \frac{∂t}{∂z_{i_{3}}} · ∑_{i_{2}=1}^{n} \frac{∂t}{∂\overline{z}_{i_{2}}}

+ e^{−2a_{i_{1}}t} ∑_{i_{3}=1}^{n} ∑_{i_{2}=1}^{n} \frac{∂^{2}t}{∂z_{i_{3}}∂\overline{z}_{i_{2}}} )dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1}=1}^{n}(−\frac{2}{a_{i_{1}}} )( ∑_{i_{2}≠i_{3}} \frac{−2a_{i_{1}}e^{−2(a_{i_{1}}+a_{i_{3}}+a_{i_{2}})t}\overline{z}_{i_{3}}z_{i_{2}}}{4M^{2}}

+ ∑_{i_{2}≠i_{3}} \frac{e^{−2a_{i_{1}}t}}{2M^{2}} ((\frac{N}{M} − (a_{i_{2}} + a_{i_{3}}))e^{−2(a_{i_{2}}+a_{i_{3}})t}z_{i_{2}} \overline{z}_{i_{3}}))dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1}=1}^{n}(−\frac{2}{a_{i_{1}}} )(∑_{i_{3}=1}^{n} \frac{−2a_{i_{1}}e^{−2(a_{i_{1}}+2a_{i_{3}})t}|z_{i_{3}}|^{2}}{4M^{2}}

+ ∑_{i_{3}=1}^{n} \frac{e^{−2a_{i_{1}}t}}{2M^{2}} (e^{−2a_{i_{3}}t}M + (\frac{N}{M} − 2a_{i_{3}})e^{−4a_{k}t}|z_{k}|^{2}))dz_{i_{3}} ∧ d\overline{z}_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{3})} \frac{1}{a_{i_{1}}} (\frac{a_{i_{1}}e^{−2(a_{i_{1}}+a_{i_{3}}+a_{i_{2}})t}\overline{z}_{i_{3}}z_{i_{2}}}{M^{2}}

− \frac{1}{M^{2}} (\frac{N}{M} − (a_{i_{2}} + a_{i_{3}}))e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}z_{i_{2}} \overline{z}_{i_{3}})dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{3}} \frac{1}{a_{i_{1}}} (\frac{a_{i_{1}}e^{−2(a_{i_{1}}+2a_{i_{3}})t}|z_{i_{3}}|^{2}}{M^{2}}

− \frac{1}{M^{2}} (e^{−2(a_{i_{1}}+a_{i_{3}})t}M + (\frac{N}{M} − 2a_{i_{3}})e^{−2(a_{i_{1}}+2a_{i_{3}})t}|z_{i_{3}}|^{2}))dz_{i_{3}} ∧ d\overline{z}_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{3}} \frac{1}{a_{i_{1}}} (−\frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{M} + \frac{1}{M^{2}} ((a_{i_{1}} + 2a_{i_{3}}) − \frac{N}{M} )e^{−2(a_{i_{1}}+2a_{i_{3}})t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ d\overline{z}_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{3})} \frac{1}{a_{i_{1}}} (\frac{1}{M^{2}} ((a_{i_{1}} + a_{i_{2}} + a_{i_{3}}) − \frac{N}{M})e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}z_{i_{2}} \overline{z}_{i_{3}})dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{3}} \frac{1}{a_{i_{1}}M^{3}} \{e^{−2(a_{i_{1}}+a_{i_{3}})t}M^{2} (21)
+ (N − (a_{i_{1}} + 2a_{i_{3}})M)e^{−2(a_{i_{1}}+2a_{i_{3}})t}|z_{i_{3}}|^{2}\}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{3})} \frac{1}{a_{i_{1}}M^{3}} \{(N − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})M)e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}z_{i_{2}} \overline{z}_{i_{3}}\}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}.

We shall calculate (21) as follows:

(♢) = ∑_{i_{1},i_{3}} \frac{1}{a_{i_{1}}M^{3}} \{e^{−2(a_{i_{1}}+a_{i_{3}})t}M^{2}

+(N − (a_{i_{1}} + 2a_{i_{3}})M)e^{−2(a_{i_{1}}+2a_{i_{3}})t}|z_{i_{3}}|^{2}\}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}.

(♠) = ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{3})} \frac{1}{a_{i_{1}}M^{3}} \{(N − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})M)e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}z_{i_{2}} \overline{z}_{i_{3}}\}
dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}.

We calculate (♢) and (♠).

(♢) = ∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (M^{2} + (N − (a_{i_{1}} + 2a_{i_{3}})M)e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}
= ∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (∑_{i_{2}=1}^{n} a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} · M
+(N − (a_{i_{1}} + 2a_{i_{3}})M)e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}
= ∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} ∑_{i_{4}=1}^{n}(∑_{i_{2}=1}^{n} a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} · a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2}

+(a_{i_{4}}^{2}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} − (a_{i_{1}} + 2a_{i_{3}})a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} [ ∑_{i_{4}≠i_{1},i_{3}}(∑_{i_{2}=1}^{n}(a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} · a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2})

+(a_{i_{4}}^{2}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} − (a_{i_{1}} + 2a_{i_{3}})a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})

+ ∑_{i_{2}=1}^{n}(a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} · a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})

+(a_{i_{3}}^{2}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} − (a_{i_{1}} + 2a_{i_{3}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}

+ ∑_{i_{2}=1}^{n}(a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} · a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})

+(a_{i_{1}}^{2}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} − (a_{i_{1}} + 2a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{3},i_{4}(i_{4}≠i_{1},i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (∑_{i_{2}=1}^{n}(a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} · a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2})

+a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2}(a_{i_{4}} − a_{i_{1}} − 2a_{i_{3}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (∑_{i_{2}=1}^{n}(a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−2a_{i_{3}}a_{i_{1}}e^{−2(a_{i_{1}}+a_{i_{3}})t}|z_{i_{1}}|^{2}|z_{i_{3}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{3}}e^{−4a_{i_{3}}t}|z_{i_{3}}|^{4})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}.

The latter half of the above is

∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (∑_{i_{2}=1}^{n}(a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−2a_{i_{3}}a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{3}}e^{−4a_{i_{3}}t}|z_{i_{3}}|^{4})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} \{  ∑ ; i_{1},i_{2},i_{3}(i_{2}≠i_{1},i_{3})(a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}
+(a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2}
+(a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}
−2a_{i_{3}}a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{3}}e^{−4a_{i_{3}}t}|z_{i_{3}}|^{4}}
dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (  ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{1},i_{3})}(a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

+a_{i_{1}}(a_{i_{1}}e^{−4a_{i_{1}}t}|z_{i_{1}}|^{4} − a_{i_{3}}e^{−4a_{i_{3}}t}|z_{i_{3}}|^{4}))dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{1},i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{M^{3}} (a_{i_{1}}e^{−4a_{i_{1}}t}|z_{i_{1}}|^{4} − a_{i_{3}}e^{−4a_{i_{3}}t}|z_{i_{3}}|^{4})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}.

Here,

∑_{i_{1},i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{M^{3}} (a_{i_{1}}e^{−4a_{i_{1}}t}|z_{i_{1}}|^{4} − a_{i_{3}}e^{−4a_{i_{3}}t}|z_{i_{3}}|^{4})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1}>i_{3}} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{M^{3}} (a_{i_{1}}e^{−4a_{i_{1}}t}|z_{i_{1}}|^{4} − a_{i_{3}}e^{−4a_{i_{3}}t}|z_{i_{3}}|^{4}
+a_{i_{3}}e^{−4a_{i_{3}}t}|z_{i_{3}}|^{4} − a_{i_{1}}e^{−4a_{i_{1}}t}|z_{i_{1}}|^{4})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}
= 0.

So,

(♢) = ∑_{i_{1},i_{3},i_{4}(i_{4}≠i_{1},i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (∑_{i_{2}=1}^{n}(a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} · a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2})
+a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2}(a_{i_{4}} − a_{i_{1}} − 2a_{i_{3}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{1},i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}.

Replace i_{2} with i_{4} for above (♢), and then,

(♢) = ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{1},i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (∑_{i_{4}=1}^{n}(a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} · a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2})

+a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}(a_{i_{2}} − a_{i_{1}} − 2a_{i_{3}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{1},i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{1},i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} [ ∑_{i_{4}(i_{4}≠i_{1},i_{2},i_{3})}(a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} · a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2})

+a_{i_{1}}a_{i_{2}}e^{−2(a_{i_{1}}+a_{i_{2}})t}|z_{i_{1}}|^{2}|z_{i_{2}}|^{2} + a_{i_{2}}^{2}e^{−4a_{i_{2}}t}|z_{i_{2}}|^{4} +a_{i_{3}}a_{i_{2}}e^{−2(a_{i_{3}}+a_{i_{2}})t}|z_{i_{3}}|^{2}|z_{i_{2}}|^{2} + a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}(a_{i_{2}} − a_{i_{1}} − 2a_{i_{3}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}]

dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3}(i_{2}≠i_{1},i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} · a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} [a_{i_{1}}a_{i_{2}}e^{−2(a_{i_{1}}+a_{i_{2}})t}|z_{i_{1}}|^{2}|z_{i_{2}}|^{2}

+a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}(a_{i_{2}} − a_{i_{1}} − a_{i_{3}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{2}}^{2}e^{−4a_{i_{2}}t}|a_{i_{2}}|^{4} +(a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (a_{i_{2}}a_{i_{4}}e^{−2(a_{i_{2}}+a_{i_{4}})t}|z_{i_{2}}|^{2}|z_{i_{4}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} a_{i_{2}}|z_{i_{2}}|^{2}[2a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} + (a_{i_{2}} − a_{i_{1}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}

+a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}.

Next, we calculate (♠).

(♠) = ∑_{i_{1},i_{2},i_{3},(i_{2}≠i_{3})} \frac{1}{a_{i_{1}}M^{3}} [(N − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})M)e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}z_{i_{2}} \overline{z}_{i_{3}}]

dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3},(i_{2}≠i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} z_{i_{2}} \overline{z}_{i_{3}}

× ∑_{i_{4}=1}^{n}[a_{i_{4}}^{2}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} z_{i_{2}} \overline{z}_{i_{3}}

×[a_{i_{4}}^{2}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3},(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} z_{i_{2}} \overline{z}_{i_{3}}[(a_{i_{1}}^{2}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})

+(a_{i_{2}}^{2}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2})

+(a_{i_{3}}^{2}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} z_{i_{2}} \overline{z}_{i_{3}}

×[a_{i_{4}}^{2}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3},(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} z_{i_{2}} \overline{z}_{i_{3}}

×[−(a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−(a_{i_{1}} + a_{i_{2}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}.

Therefore,

∂\overline{∂}ω = (♢) + (♠)

= ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} (a_{i_{2}}a_{i_{4}}e^{−2(a_{i_{2}}+a_{i_{4}})t}|z_{i_{2}}|^{2}|z_{i_{4}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{2},i_{1},i_{3},(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} a_{i_{2}}|z_{i_{2}}|^{2}

×[2a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} + (a_{i_{2}} − a_{i_{1}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} z_{i_{2}} \overline{z}_{i_{3}}

×[a_{i_{4}}^{2}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{4}}e^{−2a_{i_{4}}t}|z_{i_{4}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+ ∑_{i_{1},i_{2},i_{3},(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} z_{i_{2}} \overline{z}_{i_{3}}

×[−(a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−(a_{i_{1}} + a_{i_{2}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

= ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}+a_{i_{4}})t}}{a_{i_{1}}M^{3}} [(a_{i_{2}}a_{i_{4}}|z_{i_{2}}|^{2}|z_{i_{4}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+z_{i_{2}} \overline{z}_{i_{3}}(a_{i_{4}}^{2}|z_{i_{4}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{4}}|z_{i_{4}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}]

+ ∑_{i_{1},i_{2},i_{3},(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} [a_{i_{2}}|z_{i_{2}}|^{2}[2a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} + (a_{i_{2}} − a_{i_{1}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}

+a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+z_{i_{2}} \overline{z}_{i_{3}}[−(a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−(a_{i_{1}} + a_{i_{2}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}].

We shall consider above calculation separately for (♣) and (♡) as follows:

(♣) = ∑_{i_{1},i_{2},i_{3},(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}M^{3}} [a_{i_{2}}|z_{i_{2}}|^{2}[2a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} + (a_{i_{2}} − a_{i_{1}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}
+a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}
+z_{i_{2}} \overline{z}_{i_{3}}[−(a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−(a_{i_{1}} + a_{i_{2}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}]dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}],

(♡) = ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}+a_{i_{4}})t}}{a_{i_{1}}M^{3}} [(a_{i_{2}}a_{i_{4}}|z_{i_{2}}|^{2}|z_{i_{4}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+z_{i_{2}} \overline{z}_{i_{3}}(a_{i_{4}}^{2}|z_{i_{4}}|^{2} − (a_{i_{1}} + a_{i_{2}} + a_{i_{3}})a_{i_{4}}|z_{i_{4}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}].

So,

(♣) = ∑_{i_{1},i_{2},i_{3},(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}M^{3}} a_{i_{2}}a_{i_{3}}[a_{i_{2}}|z_{i_{2}}|^{2}(2a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} + (a_{i_{2}} − a_{i_{1}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}

+a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+z_{i_{2}} \overline{z}_{i_{3}}(−(a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−(a_{i_{1}} + a_{i_{2}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}]

= ∑_{i_{1},i_{2},i_{3}(i_{1}>i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}M^{3}}

×[a_{i_{2}}a_{i_{3}}a_{i_{2}}|z_{i_{2}}|^{2}(2a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} + (a_{i_{2}} − a_{i_{1}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}

+a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{1}}a_{i_{2}}|z_{i_{2}}|^{2}(2a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + (a_{i_{2}} − a_{i_{3}})e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2}

+a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{3}}z_{i_{2}} \overline{z}_{i_{3}}(−(a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2} − (a_{i_{1}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−(a_{i_{1}} + a_{i_{2}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2})dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{1}}z_{i_{2}} \overline{z}_{i_{1}}(−(a_{i_{2}} + a_{i_{1}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} − (a_{i_{3}} + a_{i_{1}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2}

−(a_{i_{3}} + a_{i_{2}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2})dz_{i_{1}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{3}}]

= ∑_{i_{1},i_{2},i_{3}(i_{1}>i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}M^{3}} [(a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2}

+(a_{i_{1}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} + (a_{i_{1}} + a_{i_{2}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}]

×(a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}} − a_{i_{2}}a_{i_{3}}z_{i_{2}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

−a_{i_{2}}a_{i_{1}}z_{i_{2}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{3}})

= ∑_{i_{1},i_{2},i_{3}(i_{1}>i_{3})} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}M^{3}} [(a_{i_{2}} + a_{i_{3}})a_{i_{1}}e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2}

+(a_{i_{1}} + a_{i_{3}})a_{i_{2}}e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} + (a_{i_{1}} + a_{i_{2}})a_{i_{3}}e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2}]

×a_{i_{2}}z_{i_{2}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ (a_{i_{2}} \overline{z}_{i_{2}}d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}} + a_{i_{3}} \overline{z}_{i_{3}}d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{2}} + a_{i_{1}} \overline{z}_{i_{1}}d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{3}}).

Also,

(♡) = ∑_{i_{1},i_{2},i_{3},i_{4}(≠)} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}+a_{i_{4}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}a_{i_{4}}M^{3}}

×[a_{i_{4}}^{2}|z_{i_{4}}|^{2}(a_{i_{3}}a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{3}}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})z_{i_{2}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}})]

= ∑_{i_{1}>i_{2}>i_{3}>i_{4}} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}+a_{i_{4}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}a_{i_{4}}M^{3}}

×[a_{i_{4}}^{2}|z_{i_{4}}|^{2}(a_{i_{3}}a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{3}}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})z_{i_{2}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+a_{i_{1}}a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{1}}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})z_{i_{2}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{3}}

+a_{i_{3}}a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}}

+a_{i_{1}}a_{i_{3}}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})z_{i_{1}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{2}}

+a_{i_{2}}a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}}

+a_{i_{1}}a_{i_{2}}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})z_{i_{1}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{3}}

+a_{i_{2}}a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+a_{i_{3}}a_{i_{2}}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})z_{i_{3}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{1}}a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+a_{i_{3}}a_{i_{1}}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})z_{i_{3}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}})

+a_{i_{1}}^{2}|z_{i_{1}}|^{2}(a_{i_{3}}a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}}

+a_{i_{2}}a_{i_{3}}(a_{i_{1}} − a_{i_{4}} − a_{i_{2}} − a_{i_{3}})z_{i_{2}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}}

+a_{i_{4}}a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}}

+a_{i_{2}}a_{i_{4}}(a_{i_{1}} − a_{i_{4}} − a_{i_{2}} − a_{i_{3}})z_{i_{2}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{3}}

+a_{i_{3}}a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}}

+a_{i_{4}}a_{i_{3}}(a_{i_{1}} − a_{i_{4}} − a_{i_{2}} − a_{i_{3}})z_{i_{4}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}}

+a_{i_{2}}a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}}

+a_{i_{4}}a_{i_{2}}(a_{i_{1}} − a_{i_{4}} − a_{i_{2}} − a_{i_{3}})z_{i_{4}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{3}}

+a_{i_{2}}a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{2}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}}

+a_{i_{3}}a_{i_{2}}(a_{i_{1}} − a_{i_{4}} − a_{i_{2}} − a_{i_{3}})z_{i_{3}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}}

+a_{i_{4}}a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{2}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}} +a_{i_{3}}a_{i_{4}}(a_{i_{1}} − a_{i_{4}} − a_{i_{2}} − a_{i_{3}})z_{i_{3}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}})

+a_{i_{2}}^{2}|z_{i_{2}}|^{2}(a_{i_{3}}a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{4}}a_{i_{3}}(a_{i_{2}} − a_{i_{1}} − a_{i_{4}} − a_{i_{3}})z_{i_{4}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}}

+a_{i_{1}}a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{4}}a_{i_{1}}(a_{i_{2}} − a_{i_{1}} − a_{i_{4}} − a_{i_{3}})z_{i_{4}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{3}}

+a_{i_{3}}a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}}

+a_{i_{1}}a_{i_{3}}(a_{i_{2}} − a_{i_{1}} − a_{i_{4}} − a_{i_{3}})z_{i_{1}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{4}}

+a_{i_{4}}a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}}

+a_{i_{1}}a_{i_{4}}(a_{i_{2}} − a_{i_{1}} − a_{i_{4}} − a_{i_{3}})z_{i_{1}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{3}}

+a_{i_{4}}a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}}

+a_{i_{3}}a_{i_{4}}(a_{i_{2}} − a_{i_{1}} − a_{i_{4}} − a_{i_{3}})z_{i_{3}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{1}}a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}} +a_{i_{3}}a_{i_{1}}(a_{i_{2}} − a_{i_{1}} − a_{i_{4}} − a_{i_{3}})z_{i_{3}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}})

+a_{i_{3}}^{2}|z_{i_{3}}|^{2}(a_{i_{4}}a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{4}}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})z_{i_{2}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+a_{i_{1}}a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{1}}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})z_{i_{2}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}}

+a_{i_{4}}a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{4}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}}

+a_{i_{1}}a_{i_{4}}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})z_{i_{1}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{2}}

+a_{i_{2}}a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{4}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}}

+a_{i_{1}}a_{i_{2}}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})z_{i_{1}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{4}}

+a_{i_{2}}a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+a_{i_{4}}a_{i_{2}}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})z_{i_{4}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}}

+a_{i_{1}}a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}}

+a_{i_{4}}a_{i_{1}}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})z_{i_{4}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}})]

= ∑_{i_{1}>i_{2}>i_{3}>i_{4}} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}+a_{i_{4}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}a_{i_{4}}M^{3}} [a_{i_{4}}^{2}|z_{i_{4}}|^{2}((a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})

×(−a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}} + a_{i_{2}}a_{i_{3}}z_{i_{2}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}} +a_{i_{2}}a_{i_{1}}z_{i_{2}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{3}} − a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}} +a_{i_{1}}a_{i_{3}}z_{i_{1}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{2}} + a_{i_{1}}a_{i_{2}}z_{i_{1}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{3}} −a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}} + a_{i_{3}}a_{i_{2}}z_{i_{3}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{3}}a_{i_{1}}z_{i_{3}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}})

+(a_{i_{4}} − a_{i_{2}})a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}} + (a_{i_{4}} − a_{i_{1}})a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}}

+(a_{i_{4}} − a_{i_{3}})a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}})

+a_{i_{1}}^{2}|z_{i_{1}}|^{2}((a_{i_{1}} − a_{i_{4}} − a_{i_{2}} − a_{i_{3}})(−a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}}

+a_{i_{2}}a_{i_{3}}z_{i_{2}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}} + a_{i_{2}}a_{i_{4}}z_{i_{2}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{3}} −a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}} + a_{i_{4}}a_{i_{3}}z_{i_{4}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}} +a_{i_{4}}a_{i_{2}}z_{i_{4}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{3}} − a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{2}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}} +a_{i_{3}}a_{i_{2}}z_{i_{3}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}} + a_{i_{3}}a_{i_{4}}z_{i_{3}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}})

+(a_{i_{1}} − a_{i_{2}})a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}} + (a_{i_{1}} − a_{i_{4}})a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{3}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}}

+(a_{i_{1}} − a_{i_{3}})a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{2}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}}r)

+a_{i_{2}}^{2}|z_{i_{2}}|^{2}((a_{i_{2}} − a_{i_{1}} − a_{i_{4}} − a_{i_{3}})(−a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}}

+a_{i_{4}}a_{i_{3}}z_{i_{4}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}} + a_{i_{4}}a_{i_{1}}z_{i_{4}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{3}} −a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}} + a_{i_{1}}a_{i_{3}}z_{i_{1}} \overline{z}_{i_{3}}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{4}} +a_{i_{1}}a_{i_{4}}z_{i_{1}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{3}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{3}} − a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}} +a_{i_{3}}a_{i_{4}}z_{i_{3}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}} + a_{i_{3}}a_{i_{1}}z_{i_{3}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}})

+(a_{i_{2}} − a_{i_{4}})a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{3}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}} + (a_{i_{2}} − a_{i_{1}})a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{3}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{4}}

+(a_{i_{2}} − a_{i_{3}})a_{i_{3}}^{2}|z_{i_{3}}|^{2}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}})

+a_{i_{3}}^{2}|z_{i_{3}}|^{2}((a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})(−a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}}

+a_{i_{2}}a_{i_{4}}z_{i_{2}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}} + a_{i_{2}}a_{i_{1}}z_{i_{2}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}} −a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{4}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}} + a_{i_{1}}a_{i_{4}}z_{i_{1}} \overline{z}_{i_{4}}dz_{i_{4}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{2}} +a_{i_{1}}a_{i_{2}}z_{i_{1}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{4}} ∧ d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{4}} − a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}} +a_{i_{4}}a_{i_{2}}z_{i_{4}} \overline{z}_{i_{2}}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}} + a_{i_{4}}a_{i_{1}}z_{i_{4}} \overline{z}_{i_{1}}dz_{i_{1}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}})

+(a_{i_{3}} − a_{i_{2}})a_{i_{2}}^{2}|z_{i_{2}}|^{2}dz_{i_{4}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{1}} + (a_{i_{3}} − a_{i_{1}})a_{i_{1}}^{2}|z_{i_{1}}|^{2}dz_{i_{4}} ∧ dz_{i_{2}} ∧ d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}}

+(a_{i_{3}} − a_{i_{4}})a_{i_{4}}^{2}|z_{i_{4}}|^{2}dz_{i_{2}} ∧ dz_{i_{1}} ∧ d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}})]

= ∑_{i_{1}>i_{2}>i_{3}>i_{4}} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}+a_{i_{4}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}a_{i_{4}}M^{3}}

×[a_{i_{4}}^{2}|z_{i_{4}}|^{2}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})(a_{i_{1}}z_{i_{1}}dz_{i_{3}} ∧ dz_{i_{2}} + a_{i_{2}}z_{i_{2}}dz_{i_{1}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{2}} ∧ dz_{i_{1}})

(a_{i_{1}} \overline{z}_{i_{1}}d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}} + a_{i_{2}} \overline{z}_{i_{2}}d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{3}} + a_{i_{3}} \overline{z}_{i_{3}}d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{1}})

+a_{i_{1}}^{2}|z_{i_{1}}|^{2}(a_{i_{1}} − a_{i_{2}} − a_{i_{3}} − a_{i_{4}})(a_{i_{2}}z_{i_{2}}dz_{i_{4}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{2}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{3}} ∧ dz_{i_{2}})

(a_{i_{2}} \overline{z}_{i_{2}}d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{3}} + a_{i_{3}} \overline{z}_{i_{3}}d\overline{z}_{i_{2}} ∧ d\overline{z}_{i_{4}} + a_{i_{4}} \overline{z}_{i_{4}}d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}})

+a_{i_{2}}^{2}|z_{i_{2}}|^{2}(a_{i_{2}} − a_{i_{1}} − a_{i_{3}} − a_{i_{4}})(a_{i_{1}}z_{i_{1}}dz_{i_{4}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{1}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{3}} ∧ dz_{i_{1}})

(a_{i_{1}} \overline{z}_{i_{1}}d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{3}} + a_{i_{3}} \overline{z}_{i_{3}}d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{4}} + a_{i_{4}} \overline{z}_{i_{4}}d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{1}})

+a_{i_{3}}^{2}|z_{i_{3}}|^{2}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})(a_{i_{1}}z_{i_{1}}dz_{i_{4}} ∧ dz_{i_{2}} + a_{i_{2}}z_{i_{2}}dz_{i_{1}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{2}} ∧ dz_{i_{1}})

(a_{i_{1}} \overline{z}_{i_{1}}d\overline{z}_{i_{4}} ∧ d\overline{z}_{i_{2}} + a_{i_{2}} \overline{z}_{i_{2}}d\overline{z}_{i_{1}} ∧ d\overline{z}_{i_{4}} + a_{i_{4}} \overline{z}_{i_{4}}d\overline{z}_{i_{3}} ∧ d\overline{z}_{i_{2}})].

Therefore,

∂ ¯∂ω = (♡) + (♣)

= \sqrt{−1} ∑_{i_{1}>i_{2}>i_{3}>i_{4}} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}}+a_{i_{4}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}a_{i_{4}}M^{3}}

×[a_{i_{4}}^{2}|z_{i_{4}}|^{2}(a_{i_{4}} − a_{i_{1}} − a_{i_{2}} − a_{i_{3}})(a_{i_{1}}z_{i_{1}}dz_{i_{3}} ∧ dz_{i_{2}} + a_{i_{2}}z_{i_{2}}dz_{i_{1}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{2}} ∧ dz_{i_{1}})

(a_{i_{1}} z̄_{i_{1}}dz̄_{i_{3}} ∧ dz̄_{i_{2}} + a_{i_{2}} z̄_{i_{2}}dz̄_{i_{1}} ∧ dz̄_{i_{3}} + a_{i_{3}} z̄_{i_{3}}dz̄_{i_{2}} ∧ dz̄_{i_{1}})

+a_{i_{1}}^{2}|z_{i_{1}}|^{2}(a_{i_{1}} − a_{i_{2}} − a_{i_{3}} − a_{i_{4}})(a_{i_{2}}z_{i_{2}}dz_{i_{4}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{2}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{3}} ∧ dz_{i_{2}})

(a_{i_{2}} z̄_{i_{2}}dz̄_{i_{4}} ∧ dz̄_{i_{3}} + a_{i_{3}} z̄_{i_{3}}dz̄_{i_{2}} ∧ dz̄_{i_{4}} + a_{i_{4}} z̄_{i_{4}}dz̄_{i_{3}} ∧ dz̄_{i_{2}})

+a_{i_{2}}^{2}|z_{i_{2}}|^{2}(a_{i_{2}} − a_{i_{1}} − a_{i_{3}} − a_{i_{4}})(a_{i_{1}}z_{i_{1}}dz_{i_{4}} ∧ dz_{i_{3}} + a_{i_{3}}z_{i_{3}}dz_{i_{1}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{3}} ∧ dz_{i_{1}})

(a_{i_{1}} z̄_{i_{1}}dz̄_{i_{4}} ∧ dz̄_{i_{3}} + a_{i_{3}} z̄_{i_{3}}dz̄_{i_{1}} ∧ dz̄_{i_{4}} + a_{i_{4}} z̄_{i_{4}}dz̄_{i_{3}} ∧ dz̄_{i_{1}})

+a_{i_{3}}^{2}|z_{i_{3}}|^{2}(a_{i_{3}} − a_{i_{1}} − a_{i_{2}} − a_{i_{4}})(a_{i_{1}}z_{i_{1}}dz_{i_{4}} ∧ dz_{i_{2}} + a_{i_{2}}z_{i_{2}}dz_{i_{1}} ∧ dz_{i_{4}} + a_{i_{4}}z_{i_{4}}dz_{i_{2}} ∧ dz_{i_{1}})

(a_{i_{1}} z̄_{i_{1}}dz̄_{i_{4}} ∧ dz̄_{i_{2}} + a_{i_{2}} z̄_{i_{2}}dz̄_{i_{1}} ∧ dz̄_{i_{4}} + a_{i_{4}} z̄_{i_{4}}dz̄_{i_{3}} ∧ dz̄_{i_{2}})]

+\sqrt{−1} ∑_{i_{1},i_{2},i_{3}(i_{3}>i_{1})} \frac{e^{−2(a_{i_{1}}+a_{i_{2}}+a_{i_{3}})t}}{a_{i_{1}}a_{i_{2}}a_{i_{3}}M^{3}}

×[(a_{i_{3}}(a_{i_{1}} + a_{i_{2}})e^{−2a_{i_{3}}t}|z_{i_{3}}|^{2} + a_{i_{2}}(a_{i_{1}} + a_{i_{3}})e^{−2a_{i_{2}}t}|z_{i_{2}}|^{2} + a_{i_{1}}(a_{i_{2}} + a_{i_{3}})e^{−2a_{i_{1}}t}|z_{i_{1}}|^{2}) × a_{i_{2}}z_{i_{2}}dz_{i_{3}} ∧ dz_{i_{1}} ∧ (a_{i_{2}} z̄_{i_{2}}dz̄_{i_{3}} ∧ dz̄_{i_{1}} + a_{i_{1}} z̄_{i_{1}}dz̄_{i_{2}} ∧ dz̄_{i_{3}} + a_{i_{3}} z̄_{i_{3}}dz̄_{i_{1}} ∧ dz̄_{i_{2}})].

Appendix B

In Appendix B, we shall show ∂ ¯∂ω = 0 for ω in Theorem 4.2. This proves that ω in Theorem 4.2 is a Gauduchon metric form. The calculation is similar to that of Theorem 4.1. Recall here the relation (16), which is

\frac{1}{b − 4}|α|^{2}Q^{2}|α^{t}z + tα^{t−1}w|^{2} + |α^{t}w|^{2} = 1.

Put

B := \frac{1}{b − 4}|α|^{2}Q^{2} (> 0),
M′ := (B|z|^{2} + |w|^{2})Q + B(ᾱ^{−1}zw̄ + α^{−1}wz̄) + tB(ᾱ^{−1}zw̄ + α^{−1}wz̄)Q
+2tB|α|^{−2}|w|^{2} + t^{2}B|α|^{−2}|w|^{2}Q.

Then, we have

( A ) \{ \begin{matrix}\frac{∂t}{∂z} = −\frac{B(z̄ + tᾱ^{−1}w̄)}{M′} \\ \frac{∂t}{∂\overline{z}} = −\frac{B(z + tα^{−1}w)}{M′} \\ \frac{∂t}{∂w} = −\frac{tBα^{−1}z̄ + t^{2}B|α|^{−2}w̄ + w̄}{M′} \\ \frac{∂t}{∂\overline{w}} = −\frac{tBᾱ^{−1}z + t^{2}B|α|^{−2}w + w}{M′} . \end{matrix}

The quantities \frac{∂^{2}t}{∂z∂z̄}, \frac{∂^{2}t}{∂w∂z̄}, \frac{∂^{2}t}{∂z∂w̄}, \frac{∂^{2}t}{∂w∂w̄} are given by

( B ) \{ \begin{matrix}\frac{∂^{2}t}{∂z∂z̄} = \frac{(\frac{∂}{∂z} M′) B(z + tα^{−1}w) − BM′ + B^{2}α^{−1}w(z̄ + tᾱ^{−1}w̄)}{M′^{2}} \\ \frac{∂^{2}t}{∂w∂z̄} = \frac{(\frac{∂}{∂w}M′) B(z + tα^{−1}w) − Btα^{−1}M′ + Bα^{−1}w(tBα^{−1}z̄ + Bt^{2}|α|^{−2}w̄ + w̄)}{M′^{2}} \\ \frac{∂^{2}t}{∂z∂w̄} = \frac{(\frac{∂}{∂z} M′) (tBαα^{−1}z + t^{2}B|α|^{−2}w + w) − Btᾱ^{−1}M′}{M′^{2}} \\ +\frac{(Bᾱ^{−1}z + 2tB|α|^{−2}w)(Bz̄ + Btᾱ^{−1}w̄)}{M′^{2}} \\ \frac{∂^{2}t}{∂w∂w̄} = \frac{(\frac{∂}{∂w}M′) (tBαα^{−1}z + t^{2}B|α|^{−2}w + w) − (Bt^{2}|α|^{−2} + 1)M′}{M′^{2}} \\ +\frac{(Bᾱ^{−1}z + 2tB|α|^{−2}w)(Btα−1z̄ + Bt^{2}|α|^{−2}w̄ + w̄)}{M′^{2}}, \end{matrix}

where

(\frac{∂}{∂z} M′) = Bz̄Q + Bᾱ^{−1}w̄ + tBᾱ^{−1}w̄Q

+(B(ᾱ^{−1}zw̄ + α^{−1}wz̄)Q + 2B|α|^{−2}|w|^{2} + 2tB|α|^{−2}|w|^{2}Q)\frac{∂t}{∂z} ,

(\frac{∂}{∂w} M′) = w̄Q + Bα^{−1}z̄ + tBα^{−1}z̄Q + 2tB|α|^{−2}w̄ + t^{2}B|α|^{−2}w̄Q

+(B(ᾱ^{−1}zw̄ + α^{−1}wz̄)Q + 2B|α|^{−2}|w|^{2} + 2tB|α|^{−2}|w|^{2}Q)\frac{∂t}{∂w} .

Now we carry out the calculation of ∂ ¯∂ω.

¯∂ω = \sqrt{−1}(a|α|^{2t}Q\frac{∂t}{∂w̄} dw̄ ∧ dz ∧ dz̄ + (tQ − 1)aᾱ^{−1}|α|^{2t} \frac{∂t}{∂z̄}dz̄ ∧ dz ∧ dw̄

+(tQ − 1)aα^{−1}|α|^{2t} \frac{∂t}{∂w̄} dw̄ ∧ dw ∧ dz̄ + a(t^{2}Q − 2t + \frac{b − 4}{Q})|α|^{2(t−1)} \frac{∂t}{∂z̄}dz̄ ∧ dw ∧ dw̄).

∂ ¯∂ω = \sqrt{−1}\{a|α|^{2t}Q (Q\frac{∂t}{∂w} \frac{∂t}{∂w̄} + \frac{∂^{2}t}{∂w∂w̄} ) dw ∧ dw̄ ∧ dz ∧ dz̄

+aᾱ^{−1}|α|^{2t} (tQ^{2} \frac{∂t}{∂w} \frac{∂t}{∂z̄} + (tQ − 1)\frac{∂^{2}t}{∂w∂z̄}) dw ∧ dz̄ ∧ dz ∧ dw̄

+aα^{−1}|α|^{2t} (tQ^{2} \frac{∂t}{∂z} \frac{∂t}{∂w̄} + (tQ − 1)\frac{∂^{2}t}{∂z∂w̄} ) dz ∧ dw̄ ∧ dw ∧ dz̄

+aᾱ^{−1}|α|^{2(t−1)}((t^{2}Q^{2} + (b − 6)) \frac{∂t}{∂z} \frac{∂t}{∂z̄}

+(t^{2}Q − 2t + \frac{(b − 4)}{Q})\frac{∂^{2}t}{∂z∂z̄})dz ∧ dz̄ ∧ dw ∧ dw̄\}

= −\sqrt{−1}a|α|^{2t}\{(Q^{2} \frac{∂t}{∂w} \frac{∂t}{∂w̄} + Q\frac{∂^{2}t}{∂w∂w̄} ) − ᾱ^{−1} (tQ^{2} \frac{∂t}{∂w} \frac{∂t}{∂z̄} + (tQ − 1)\frac{∂^{2}t}{∂w∂z̄})

−α^{−1} (tQ^{2} \frac{∂t}{∂z} \frac{∂t}{∂w̄} + (tQ − 1)\frac{∂^{2}t}{∂z∂w̄} )

+|α|^{−2}((t^{2}Q^{2} + (b − 6))\frac{∂t}{∂z} \frac{∂t}{∂z̄}

+(t^{2}Q − 2t + \frac{(b − 4)}{Q})\frac{∂^{2}t}{∂z∂z̄})\}dz ∧ dw ∧ dz̄ ∧ dw̄

= −\sqrt{−1}a|α|^{2t}\{Q^{2} \frac{∂t}{∂w} \frac{∂t}{∂w̄} − ᾱ^{−1}tQ^{2} \frac{∂t}{∂w} \frac{∂t}{∂z̄} − α^{−1}tQ^{2} \frac{∂t}{∂z} \frac{∂t}{∂w̄} + |α|^{−2} (t^{2}Q^{2} + (b − 6)) \frac{∂t}{∂z} \frac{∂t}{∂z̄}

+Q\frac{∂^{2}t}{∂w∂w̄} − α^{−1}(tQ − 1)\frac{∂^{2}t}{∂z∂w̄} − ᾱ^{−1}(tQ − 1)\frac{∂^{2}t}{∂w∂z̄}

+|α|^{−2} (t^{2}Q − 2t + \frac{(b − 4)}{Q}) \frac{∂^{2}t}{∂z∂z̄}\}dz ∧ dw ∧ dz̄ ∧ dw̄

= −\sqrt{−1}a|α|^{2t}\{Q^{2} (\frac{∂t}{∂w} − tα^{−1} \frac{∂t}{∂z̄}) (\frac{∂t}{∂w̄} − tᾱ^{−1} \frac{∂t}{∂z̄}) + |α|^{−2}(b − 6)\frac{∂t}{∂z} \frac{∂t}{∂z̄}

+Q\frac{∂^{2}t}{∂w∂w̄} − α^{−1}(tQ − 1)\frac{∂^{2}t}{∂z∂w̄} − ᾱ^{−1}(tQ − 1)\frac{∂^{2}t}{∂w∂z̄}

+|α|^{−2} (t^{2}Q − 2t + \frac{(b − 4)}{Q}) \frac{∂^{2}t}{∂z∂z̄}\}dz ∧ dw ∧ dz̄ ∧ dw̄

= (ℵ).

Using (A) and (B), we rewrite (ℵ). Then, we have

(ℵ) = −\frac{\sqrt{−1}a|α|^{2t}}{M′^{2}} [Q^{2}|w|^{2} − 2QM′

+(tBα^{−1}z̄ + t^{2}B|α|^{−2}w̄ + w̄)(BQᾱ^{−1}z + tBQ|α|^{−2}w + B|α|^{−2}w)

+(Bz̄ + tBᾱ^{−1}w̄)((b − 5)B|α|^{−2}z + t(b − 6)Bα^{−1}|α|^{−2}w + Qα^{−1}w

−tQB|α|^{−2}z − t^{2}QBα^{−1}|α|^{−2}w)

+ (\frac{∂}{∂w} M′) (Qw + Bᾱ^{−1}z + tB|α|^{−2}w)

+ (\frac{∂}{∂z} M′) (Qz − tB|α|^{−2}z + α^{−1}w − t^{2}Bα^{−1}|α|^{−2}w)]dz ∧ dw ∧ dz̄ ∧ dw̄

= −\frac{\sqrt{−1}a|α|^{2t}}{M′^{2}} [Q^{2}|w|^{2} − 2QM′ + Bᾱ^{−1}(t^{2}B|α|^{−2}w̄ + tBα^{−1}z̄ + w̄)(tQα^{−1}w + Qz + α^{−1}w)

+B(tᾱ^{−1}w̄ + z̄)(−t^{2}QBα^{−1}|α|^{−2}w − tQB|α|^{−2}z + t(b − 6)Bα^{−1}|α|^{−2}w

+Qα^{−1}w + (b − 5)B|α|^{−2}z)

+(Qw + Bᾱ^{−1}z + tB|α|^{−2}w)(w̄Q + Bα^{−1}z̄ + tBQα^{−1}z̄ + 2tB|α|^{−2}w̄ + t^{2}BQ|α|^{−2}w̄)

+(Qz − tB|α|^{−2}z + α^{−1}w − t^{2}Bα^{−1}|α|^{−2}w)(BQz̄ + Bᾱ^{−1}w̄ + tBQᾱ^{−1}w̄)

−(BQ(ᾱ^{−1}zw̄ + α^{−1}wz̄) + 2B|α|^{−2}|w|^{2} + 2tBq|α|^{−2}|w|^{2})]dz ∧ dw ∧ dz̄ ∧ dw̄

= −\frac{\sqrt{−1}a|α|^{2t}}{M′^{2}} [t^{3}(B^{2}Q|α|^{−4}|w|^{2} − B^{2}Q|α|^{−4}|w|^{2} + B^{2}Q|α|^{−4}|w|^{2} − B^{2}Q|α|^{−4}|w|^{2})

+t^{2}(−2BQ^{2}|α|^{−2}|w|^{2} + B^{2}Qᾱ^{−1}|α|^{−2}zw̄ + B^{2}|α|^{−4}|w|^{2} + B^{2}Qα^{−1}|α|^{−2}wz̄ +(b − 6)B^{2}|α|^{−4}|w|^{2} − B^{2}Qᾱ^{−1}|α|^{−2}zw̄ − B^{2}Qα^{−1}|α|^{−2}wz̄ +B^{2}Qα^{−1}|α|^{−2}wz̄ + 2B^{2}|α|^{−4}|w|^{2} + B^{2}Qᾱ^{−1}|α|^{−2}zw̄ + B^{2}Q^{2}|α|^{−2}|w|^{2} −B^{2}Qα^{−1}|α|^{−2}wz̄ − B^{2}|α|^{−4}|w|^{2} − B^{2}Qᾱ^{−1}|α|^{−2}zw̄)

+t(−2BQ^{2}(ᾱ^{−1}zw̄ + α^{−1}wz̄) − 4BQ|α|^{−2}|w|^{2} + B^{2}Q|α|^{−2}|z|^{2} + B^{2}α^{−1}|α|^{−2}wz̄ +BQ|α|^{−2}|w|^{2} + B^{2}(b − 5)ᾱ^{−1}|α|^{−2}zw̄ + BQ|α|^{−2}|w|^{2} +(b − 6)B^{2}α^{−1}|α|^{−2}wz̄ − B^{2}Q|α|^{−2}|z|^{2} + BQ|α|^{−2}|w|^{2} + B^{2}α^{−1}|α|^{−2}wz̄ +BQ^{2}α^{−1}wz̄ + 2BQ|α|^{−2}|w|^{2} + B^{2}Q|α|^{−2}|z|^{2} + 2B^{2}ᾱ^{−1}|α|^{−2}zw̄

−2BQ|α|^{−2}|w|^{2} − B^{2}Q|α|^{−2}|z|^{2} − B^{2}ᾱ^{−1}|α|^{−2}zw̄ + BQ^{2}ᾱ^{−1}zw̄ + BQ|α|^{−2}|w|^{2})

+(Q^{2}|w|^{2} − 2Q^{2}(B|z|^{2} + |w|^{2}) − 2BQ(ᾱ^{−1}zw̄ + α^{−1}wz̄) + BQᾱ^{−1}zw̄ + B|α|^{−2}|w|^{2} +B^{2}(b − 5)|α|^{−2}|z|^{2} + BQα^{−1}wz̄ + Q^{2}|w|^{2} + BQα^{−1}wz̄ + BQᾱ^{−1}zw̄ +B^{2}|α|^{−2}|z|^{2} − BQᾱ^{−1}zw̄ − BQα^{−1}wz̄ − 2B|α|^{−2}|w|^{2} + BQ^{2}|z|^{2}

+BQᾱ^{−1}zw̄ + BQα^{−1}wz̄ + B|α|^{−2}|w|^{2})]dz ∧ dw ∧ dz̄ ∧ dw̄

= −\frac{\sqrt{−1}a|α|^{2t}}{M′^{2}} [t^{2}B(B(b − 4)|α|^{−2} − Q^{2})|α|^{−2}|w|^{2}
+tB(B(b − 4)|α|^{−2} − Q^{2})(ᾱ^{−1}zw̄ + α^{−1}wz̄)

+B(B(b − 4)|α|^{−2} − Q^{2})|z|^{2}]dz ∧ dw ∧ dz̄ ∧ dw̄

= 0,

since we have set B = \frac{1}{b − 4}|α|^{2}Q^{2}. Thus, we get dd^{c}ω = 0 and conclude that (17) is indeed a Gauduchon metric form.

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Reitaku Mizunami Junior & Senior High School 1661 Hagiwara Inatsu-cho, Mizunami-shi, Gifu 509-6102 JAPAN e-mail : statsumi@mz.reitaku-u.ac.jp